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TAM THaC S~C HAl A. LY THUYET CAN NHO
3. D;nh Ij Viet
a. Ph~n thuan: Ne'u phirong trlnh b~c hai co 2 nghiern Xl' X2 thi:
{s : Xi + " : - ~
P - XI,X2 -a
I. Phuong trmh b~c hai: ax2 + bx + c = 0 (1)
1. Cach giai va hi?n luijll trong truang h(lp tlJllg quat:
+ Neu a = 0 ta dU<;1c phuong trlnh b~c nha't: bx + c = O. Bien luan nhu sau:
a. Ne'u b * 0 thl phirong trlnh co nghiem duy nha't: x = - ~ .
b. Ne'u b = 0 va c = 0 thl phuong trlnh nghiern dung '\Ix E R
c. Neu b = 0 va c * 0 thl phirong trlnh vo nghiern. + Neu a * O. Tinh ~ (hoac ~')
(V di ~ = b2 - 4ac; ~. = b .2 - ac; b · = b ) 2
- Ne'u ~ < 0 thl phuong trlnh vo nghiern.
N'" A 0 hl h ' h co h' ~ ke b
- eu L.l = tIP irong trin co ng rem ep: Xo = - 2a
- Neu ~ > 0 thl phirong trlnh co hai nghiern phan biet:
-b-~ -b+~
Xl = V X2 =
2a 2a
2. Chuj
+ Neu a.c < 0 thl phirong trlnh b~c hai luon co hai nghiern phan biet, + Ne'u Xl" x2la hai nghiern cua phirong trlnh thl:
ax2 + bx + c = a(x - Xl )(X - x2)
+ Neu a + b + c = 0 thl (1) c6 hai nghiem la: Xl = 1 V X2 = ~
a
+Ne'ua-b+c=Othl(l)cohainghi~mla: Xl =-lvX2 =-~ a
5
{a.f(l3) < 0 • a. < Xl < 13 < X2 <=> f
a. (u) > 0
{a.f(a.) < 0 .XI < a. < X2 < 13 <=> a.f'(B) > 0
b. Ph~n dao: Neu hai s6 Xl' X2 thea
{Xl + X2 = S
XI,X2 = P (vdi 82 - 4P ;;:;: 0)
thl Xl' X2 la hai nghiern ciia phirong trlnh: X2 - 8x + P = 0
4. Da'u cua nghiern s6
Neu phirong trinh b~e hai: ax2 ,+ bx + c = 0 (a 'i:- 0) e6 hai nghiern Xl' X2 thl:
• Xl < 0 < X2 <=> P < 0
• Xl < X2 < 0 <=> {~ : ~ • 0 < Xl < X2 <=> {~ : ~
8<0 8>0
II. Tam thuc b{ic hai : f(x) = ax2 + bx + c (a 'i:- 0)
1. D{{u cua tam thuc
• f'(x) < 0 vdi \ix E R <=> {a < 0 ~<O
. {a> 0
• f'(x) > 0 vdi \ix ER <=>
~<O
• Ne'u ~ = 0 thl a.f(x) > 0 \ix 'i:- - _E_ 2a
• Neu ~ > 0 thl f(x) = 0 e6 hai nghiern Xl' X2' XI < X2 + Neu Xl < X < x2 thl a.f(x) < 0
+ Ne'u [X < Xl thl a.f(x) > 0
x > x2
2. Cong thirc so sanh cac s6 0.,13 vdi hai nghiern ella phuong trtnh b~e hai: (a. < P) • Xl < a. < x2 <=> a.f'(«) < 0
D;llh Ij:
• Ne'u t5n tai s6 thirc a. sao cho a.f'(n) < 0 thl phirong trlnh b~c hai
f(x) = ax2 + bx + c = 0 (a :t: 0). C6 hai nghiern Xl' X2 thea man Xl < a. < X2
~>O
• a. < Xl < X2 <=> a.f'(n) > 0 8 '--0.>0
2
~>O
• Xl < X2 < a. <=> a.f'(a) > 0
8_0.<0 2
6
Ll>O a.f'(B) > 0 • U < Xl < X2 < 13 <=> a.f'(«) > 0 S u<-<13 2
{a.f(U) < 0 • Xl < U < 13 < x2 <=>
a.f(l3) < 0
B. CAC CHUYEN DE TOAN vA PHUONG PHAP GIA.I
CHUYENDE 1:
GIA.I vA nrtN LU~N PHUONG TRINH n~c NHAT
1. Phuong phap
• Birdc 1: Bie'n d6i phirong trlnh da: cho.ve dang: ax + b = 0 (1)
• Birdc 2: Xet cac tnrong hop sau:
TH1: Ne'u a = 0; the' vao (1) va ki~m tra.
TH2: Ne'u a i:- 0 thl (1) <=> X = _ b
a
• Bu'd'c 3: Ke't luan,
2. Bili H}p
Bili 1: Giai va bien Iuan phirong trlnh: 2x + 3m = mx + 2 (1)
(1) <=> (2 - m)x = 2 - 3m (2)
• Neu m = 2 thl (2) <=> Ox = -4 vo 19 ~ (2) vo nghiern
• Ne'u m i:- 2 thl (2) <=> x = 2 - 3m 2-m
Bal 2: Giai va bien lu~n:
2x+3m m 2m-l
-::--- = -- +--
x2 -1 x+ 1 x-1
(1)
Dieu kien: x i:- ±1
(1) <=> (3m - 3)x = 2m + 1 (2)
• Neu 3m - 3 = 0 <=> m = 1 thl (2) vo nghiern
N,." 1 hl (2) 2m + 1
• eu m i:- t 1 <=> X = --
3m-3
7
{_2m+1*-1 {m*-4
f)i~u kien: x *- ±1 <=> 3m - 3 <=> 2
2m+1 1 m*--
. 3m-3 *- - 5
Ke't luan:
+ Khi m = 1; m = 4; m = ~thl (1) vo nghiern.
5 .
Khi 1 4 2 thl h trinh cc hiA d nh'" 2m + 1
+ m *- /\ m *- /\ m *- - p u'cJng co ng ern uy at x = --
5 3m-3
B"3 G'?' :'b'A 1 A h 1 h 2mx-3 x-m (1)
81 : tar va ien uan p irong tr n : LX - -_
"X - Fx
Giai .
f)i~u kien: X > 0
(1) <=> 2mx - 3 = X - m <=> (2m - 1)x = 3 - m (2)
• Ne'u m = 4 thl (2) vo nghiern
'" 1 1 2 2m-1
• Neu m *- - th ( ) X = ---
2 3-m
D, ~ ki A 0 2m - 1 0 1 3
uleu len: X > <=> > <=> - < m <
. 3-m 2
,. ;It
CHUYENDE 2:
NGHItM CVA PHU<1NG TRiNH n!c NHAT THOA MAN MOT DIEU KItN CHO TRUde
1. Phuong phap
• Biroc 1: Dira phuong trlnh da eho v~ dang: ax + b = 0 (1)
• Birdc 2: TIm di~u kien cua a d~ (1) e6 nghiem xo' Cho nghiem Xo thea man di~u kien.
2. Bai t~p
Bai 1: Cho phuong trlnh:
( 2m + 1)x - 3m + 2 = 3x + m
(1)
TIm m d~ phirong trlnh e6 nghiern x E (0; 3)
Giai
(1) <=> (2m - 2)x = 4m - 2 <=> (rn -1)x = 2m -1 (2)
• Ne'u m = 1 thl (2) vo nghiern
'" 2m-1
• Neu m *- 1 thl (2) <=> x = -m-1
8
{2m -1 > 0 '[ 1
m-l m<-
<=> <=> 2
2m -1 3 2
---< m>
m-l
Yay cac gia tri cua m dn tim lit: m <.!. v m > 2
" 2
Nghiern x E (0; 3) <=> 0 < 2m - 1 < 3 m-l
Bal 2: Cho phircng trlnh: (3m - 2)x - m = 4mx + 2m - 5
Tim m d~ phirong trtnh eo nghiern nguyen,
Giai
(1) <=> (rn + 2)x = 5 - 3m(2)
(2) 0 h'A 2 ' hie l' 5-3m 3 11
• . e ng It;m <=> m 7= - va ng It;m a x = --- = - +--
m+2. m+2
• f)~ X E Z+ thl m + 2 la' u'O'e s6 cua 11
m+2=-1 m=-3
m + 2 = -11 m = -13
<=>
<=>
m+2 =1
m+2=11
m=-1 m'=9
Bai 3: Cho phirong trlnh:
~ x-I [(2m - 3 )x + m + ( 1 - m)x - 3 J = 0 (1)
Tim m d~ phirong trlnh co 2 nghiern phan biet.
Giai
(1) ¢> [ {~:: 1_ 3)x + m + (1 _ mlx _ 3 ~ 0
<=> f{x x= >11
l (rn - 2)x = 3 - m (2)
f)~ (1) eo·2 nghiern phan biet thl (2) eo dung 1 nghiern 100n hon 1
<=> {m 7= : _ m <=> {: : 2~ <=> {m 7= 2 5 <=> 2 < m < ~
x=-->1 >0 2<m<- 2
m-2 m-2 2
Yay de gia tri cua m dn tim la: 2 < m < ~.
.. 2
Bai 4: Cho phirong trlnh:
( 2m - 1) + ( 3 - n )( x - 2) _: 2m + n + 2 = 0
Tim m va n d~ phirong trtnh eo nghiern dung \/x.
9
Giai (1) <=> (2m - n + 2)x = 2m - 3n + 4
{I
• . 2m - n + 2 = 0 m = --
Be (1) co nghiern dung vdi moi x thl: { <=> 2
2m - 3n + 4 = 0 n = 1
CHUYEN DE 3: GIAI VA Br@;N Ln~N PHUONG TRINH B~C HAl
3. Ke't luan.
1. Phu'dng phap
1. Xet a = O. Bien luan phtrong trmh b~c nhft.
2. x« a *0. Tinh ~ (ho~c ~ ')
a. Ne'u ~ < 0 thl phirong trlnh vo nghiern.
b. Ne'u ~ = 0 thl phirong trlnh co nghiern kep x = - 2~
c. Ne'u ~ > 0 thl phirong trlnh co hai nghiem phan biet.
-h-..M -h+..M
Xl = 2a v X2 = --2-a-
2. Bai t~p .
Ba! 1: Gidi va bien luan phuong trlnh:
(m-1)x2 +(2m-3)x+m+1=0 (tham sd rn) (1)
Giai
a. Ne'u rn - 1 = 0 <=> rn = 1 thl (1) tTd thanh: -x + 2 = 0 <=> x = 2
b. Ne'u m * 1. Ta co: .~ = (2m - 3)2 - 4(m -l)(m + 1) = 13 -12m
• Ne'u ~ < 0 <=> 13 -12m < 0 <=> m > ~~ thl p~u'dng trtnh (1) vo nghiern
N'" 0 13 hl (1) , hie ke 2m-3 5
• eu ~ = <=> m = _. t 1 co ng lern ep: Xo = - =
12 . 2(m -1)
• Neu ~ > 0 <=> m < 13 thl (1) co 2 nghiern phan biet:
12 ..
Xl 2 = 3 - 2m ±..M (voi ..M = '-'13 - 12m)
. . 2(m-1)
c. Ke't luan:
+ Ne'u rn = 1 thl phirong trtnh co nghiern x = 2
NA' 13 hl h ' h co hi A 5
+ eu m = 12 tIP u'dng trill co ng tern x =
NA' 13 hl h Inh A hie
+ eu rn > 12 tIP u'dng tnn vo ng iern
10
N-< 13 ht h tnh co 2 hie
+ eu rn < 12 t t p uong tnn co ng rem.
Xl2 = 3 - 2m ± Ji. v6'i Ji. = ~13 - 12m
, 2(m -1)
Hai 2: Giai va bien luan phirong trlnh:
X2 - 2(a + l)x + 2a + 5 = 0 X2 - 3x + 2
(1)
(tharn s6 a)
Giai • Dieu kien: X2 _ 3x + 2 ;I:. 0 <=> {X ;I:. 2
, x ee I
Khi do (1) trd thanh: f(x) = x2 - 2(a + l)x + 2a + 5 = 0 (2)
Ta co: /1' = (a + 1)2 - (2a + 5) = a2 - 4
+ Neu /1' < 0 <=> -2 < a < 2 thi (2) vo nghiern =>(1) cling vo nghiern.
+ Ne'u /1' = 0 <=> [a =:: 2 khi do (2) co 1 nghiern kep Xo = a + 1
a =-2
Vdi a = 2 => Xo = 3(nh~n)
Vdi a = -2 => Xo = -1(nh~n)
+ Neu /1' > 0 <=> lal > 2
• Dieu kien (2) phai co 2 nghiern khac 1 va khac 2 nen:
{f(l) ;I:. 0 {4 ;I:. 0 5
<=> <=>a;l:.-
f(2) ;I:. 0 -2a + 5 ;I:. 0 2
Khi do hai nghiern: Xl = a + 1 ± Ji.. vdi /1 = a2 - 4 Ke't luan:
+ Ne'u lal < 2 hoac a = % thi phuong trinh vo ~ghi~rn,
+ Neu a = 2 v a = -2 thi phirong trinh co nghiern kep: X = -1 v X = 3 + Ne'u lal > 2/\ a ;I:. % thi phirong trinh co 2 nghiern phan biet:
Xl 2 = a + 1 ± Ji. vdi /1 = a 2 - 4
H' . 3 G'?' >. bi A 1 A h ' h h ' b 1 a - b a + b (1)
81 : iar va len uan p irong trtn sau t eo a va : X + - = -- + --
, , X a-i b a-b
Giai
• Khi a = ±b thi (1) vo nghiern.
• Khi a ;I:. ±b , Di~u kien X ;I:. 0,
Phuong trinh (1)<=> (a2 - b2)X2 - 2(a2-+ b2)x + a2 - b2 = 0 (2)
Phuong trinh (2) co: /1' = (a2 + b2)2 - (a2 - b2)(a2 - b2) = 4a2b2 ~ 0
11
I) d' (2) , 2 h' A (a - b) (a + b) o 0 eo - ng iern: Xl = (a + b) v X2 = -(a---b-)
Khi a "1=:.1: b thl hai nghiem nay di~u khac O. Nen Xl' X2 Ia nghiem cua phuong trlnh (l). Ke't luan: + Khi a = ±b thl (1) vo nghiern.
Khi b hi (1) , 2 h' A a - b a + b
+ 1 a "1= ± t I eo ng iern: Xl = -- V X2 = --
a+b a-b
Bili 4: Giai va bien luan phirong trinh:
f(x) = mx2 + 2(2m -1)x + m = 0 vdi -1 s X~ 1 (1)
Giiii
+ Ne'u m = 0 thl (1) trCf thanh: -2x = 0 <=:> x = 0 (nhan)
, . S S
+ Neu m "1= O. Tinh ~; mf(-1);- + 1;mf(l);--1
2 2
• A' = 3m2 - 4m + 1
, 1
.~ =0<=:>m=1vm=-
3
• mf(-1) = m(2 - 2m); mf'(+I) = 0 <=:> m = 1 v m = 0
S 1-m S
• -+1=--' -+1=0<=:>m=1
2 m ' 2
1
• mf'(I) = m(6m - 2); mf(1) = 0 <=:> m = - v m = 0 , 3
S 1- 3m S 1
• --1=---' --1=0<=:>m=-
2 m' 2 ' 3
+ Ta l~p bang xet da'u sau:
**S \ S Vi tri cua nghiern
**
m ~ m(f(-I) -+1 mf(l ) --1
2 2 (XI ~ X2)
'-00
+ - - + - XI <-1 <X2< 1
0 0 0=
+ + + - + -1 < XI < 1 < X2
1 0- v Ngh ~m kep x, = 1
- r-O -0-'< r-,
3
- + + + - VO nghiern
1 -0- f-O--O .<v Nghi m kep x, =-1
r-,
+ - - + - XI < -1 < X2 < 1
+00
. 12
Bien luan: OVa vao bang xet da'u ta co ket qua:
+ Ne'u m = 1 v m = ~ thl phirong trlnh co nghiern kep.
+ Neu ~ < m < 1 thl phirong trlnh VD nghiem.
NK 0 1 hl h lnh co hi A 1- 2m + j;;::
+ eu m < v m >"3 t 1 P irong tnn co ng iern: x2 = m
1 1 - 2m - j;;::
+ Ne'u 0 < m <"3 thl plnrong trlnh co nghiern: Xl = m
Ke't luan:
+ Neu m = 0 thl phirong trlnh da cho co nghiern x = 0 + Ne'u m = 1 thl phirong trtnh da cho co nghiern x = -1
+ Neu m = ~ thl phuong trlnh da cho co nghiern x = 1
+ Neu ~ < m < 1 thl phuong trlnh da cho VQ nghiem,
1 . 1 - 2m + j;;::
+ Ne'u m < 0 v m >"3 thl phirong trtnh da cho co nghiern: x2 = m
NK 0 1 h' h Inh co hi A 1 - 2m - j;;::
+ eu < m < "3 t 1 P irong tnn co ng iern: Xl = m
CHUYENDE 4:
CHUNG MINH MOT PHUONG TRINH n!c HAl CO NGHItM 1. Phu'dng phap
Cdcli J: ca.« 2:
Cach 3:
Cach 4:
Tim s6 a. sao cho af( a. ) < 0
Tim hai s6 c; 13 sao cho f(o.).f(l3) < 0 Chung minh L\ 2: 0
Sa dung dinh 19 lien tuc:
"Cho ham s6 y = f(x) lien tuc tren mien D. Neu t6n tai a;b E 0 sao cho f(a).f(b) < 0 thl phuong trinh f(x) = 0 luon co it nha't m9t nghiern X E (a; b) "
Dung dinh ly Lagrange:
"Cho ham s6 y = f(x) lien -tuc tren [a; b] , co dao ham (6n tai tren
(a; b). khi do, t6n tai s6 a. E (a; b) sao cho: f(a)-f(b) =f'(a.Xa-b)
Cach5:
2. Bai t4p
13
Bai 1: Chung minh ding: vdi ba s6 a, b, c phan bier thl phirong trtnh:
1 1 1 0' 2 hi A \ A bi A
-- + -- + -- = co ng lym pllan IYt.
x-a x-b x-c
Giiii Phuong trinh dfl cho nrong dirong vdi:
(x - a)(x - b) + (x - b)(x - c) + (x - e) (x - a) = 0
<=> 3x2 - 2(a + b + e)x + ab + ca = 0
Ta co: ;)." = a2 + b2 + e2 - (ab + be + ea)
= i [(a - b)2 + (b - e)2 + (c - a)2 ] > 0 Va:j:. b :j:. e ~ Phu'dng trlnh dfl cho luan co hai nghiern ~ (dpcm)
--~------------~
Ba! 2: Chung minh rang: neu (a + e)2 < ab + be - 2ae thl phirong trlnh sau luon
co nghi¢m: ax2 + bx + e = 0 (1) (a :j:. 0) "
Giiii
Phirong trtnh (1) co: ;). = b2 - 4ae
Tir gia thi€t : (a + e)2 < ab + be - 2ae <=> 2(a + e)2 - 2ab - 2be + 4ae < 0
<=> (a + e)2 - 2b(a + e) + b2 + (a + e)2 + 4ae - b2 < 0 <=> b2 - 4ae > (a + e - b)2 + (a + e)2 :2: 0
<=> b2 - 4ae > 0 ~ ;). > 0
~ phuong trlnh (1) luon co nghiem (dpcm).
Bai 3: Cho a, b, c E R. Chung minh phirong trlnh sau co nghiern: x2 + 2(a + b + e)x + 3(ab + be + ea) = 0
Giiii Ta co : ;)." = a 2 + b2 + e2 - (ab + be + ea)
Chung minh: ;)'":2: 0" Th~t v~y: Theo ba't dgng thirc Cosi, tadu'<;1c: a2 + b2 :2: 2Ja2b2 = 21abl :2: 2ab
Tirong tit b2 + e2 :2: 2be; e2 + a2 :2: 2ea Cong ve' thea ve' ba ba't dhg thirc tren ta du'<;1c:
a2 + b2 + b2 + e2 + e2 + a2 :2: 2(ab + be + ea)
<=> a2 + b2 + e2 :2: ab + be + ea <=> a2 + b2 + e2 - (ab + be + ea) :2: 0 ~ ;)." :2: 0 ~ phirong trlnh dil cho luon c6 nghiern ~ dpcm,
Hai 4: Cho hai phirong trlnh: x2 + a1x+ ~ = 0
(Ir
14
va x2 + a2x + b2 = 0 (2)
Chung minh rang: Ntu a1 .a2 ~ 2(b1 + b2) thl it nha't co mQt trong hai phirong trlnh co nghiern.
Giai
Phuong trlnh (1) co: ~l = ai - 4b1
Phuong trlnh (2) co: ~2 = a~ - 4b2 => ~l + ~2 = ai + a~ - 4(b1 + b2) Theo ba't ahg thirc Cosi:
ai + a~ ~ 2Jai .a~ = 21ala21 ~ 2a1a2 ~ 2.2(b1 + b2) (theo a~ cho) => ai + a~ - 4(b1 + b2) ~ 0 => ~l + ~2 ~ 0
=> ft nha't co mot trong hai s6 ~l hoac ~2 khong am, nghia la mQt trong hai phirong trtnh aa cho luon co nghiern.
Bal 5: Chung minh r~ng: ntu m, n la hai s6 thea man: 191ml + 51nl ~ 2000 thl phuong trlnh sau co nghiern.
20mx2 + 5nx + 100 - m = 0
(1)
Giai + Vdi m = 0, tu (1) => 5n.x + 100 = 0 (*)
Theo gia thitt thl n -:t:. 0 nen phirong trlnh (*)1uon co nghiern duy nha't, nghia la (1) luon co nghiern.
+ Vdi m -:t:. 0: ta co:
~ = 25n2 - 20m(100 - m).4 = 80m2 - 8000m + 25n2 ~ 80m2 - 4(191ml + 5lnl)m + 25n2 (do gilii thiet)
~ 4m2 - 20lmnl + 25n2 = (5Inl- 21ml)2 ~ 0
=> ~ ~ 0 => Phuong trlnh (1) luon co nghiern.
Bai 6: Cho a, b, c la ba s6 khac khong, con p. q E R. Chung minh rhng:
a 2 b2 1 A 0 hi A ,
-- + -- = c uon c ng rem x -:t:. p va x -:t:. q .
x-p x-q
Giai
Phuong trlnh ail cho
<=> cx2 - [ c(p + q) + a 2 + b2J X + cpq + a 2q + b2p = 0
Ta c6: ~ = [c(p + q) + a2 + b2]2 - 4c(cpq + a2q + b2p)
= (p + q)2C2 + 2c(q + p)(b2 _ a2) + (a2 + b2)2 = (p + q)2(-4a2b2) s OVa, b -:t:. 0 va p, q E R
15
Xem ~ Ia tam thtrc b~c hai thea in s6 c thl:
{(p+Q)2 ~ 0 .
. ~ ~ ~ Ovtii Vc
~ ::;;0
Suy ra phirong trlnh da cho luon co nghiern vdi V a, b, c va p, q E R
Bai 7: Cho ba s6 a, b, c dirong thea man: a + 2b + 3c = 1
Chung minh ding: it nha't mQt trong hai phirong trlnh sau co nghiern thirc: 4x2 - 4(2a + l)x + 4a2 + 192abc + 1 = 0 (1)
4x2 - 4(2b + l)x + 4b2 + 96abc + 1 = 0 (2)
Giiii
Phuong trlnh (1) co: ~~ = 16a(1- 48bc)
Phuong trlnh (2) co: ~~ = 16b(1-- 24ac)
M~t khac: (1- 48bc) + (1- 24ac) = 2 [1-12c(2b + a)]
= 2[1-12c(1- 3c)] = 2(6c _1)2 ~ 0
Suy ra it nha't co mot trong hai s6la: (1- 48bc) hoac (1- 24ac) phai khong t-
am, suy ra co It nha't mQt s6 ~~ hoac ~~ khong am, nen luon co it nha't mQt
phirong trlnh co nghiern => (dpcrn)
Bai 8: Cho a ;t:. b ;t:. c ;t:. 0 va a, b, c E R. Chung minh r~ng phuong trlnh sau luon co nghiern: f'(x) = ab(x - a)(x - b) + bc(x -b)(x - c) + ca(x - c)(x - a) = 0
Giiii
Ta co: f'(a) = bc(a - b)(a - c); feb) = ac(b - a)(b - c) f(c) = ab(c - a)(c - b)
=> f(a).f(b).f(c) = a2b2c2(a - b)(a - c)(b - a)(b - c)(c - a)(c - b)
= _3a2b2c2(a - b)2(b - C)2(C - a)2 < 0 va, b, c, dai mQt khac nhau va khac O.
=> it nha't co met trong ba s6: f'(a); feb); f(c) phai am. Gicl sU', s6 do la: f(a) < 0
Lai co: f'(O) = a2b2 + b2c2 + c2a2 > 0 => f(a).f(O) < 0 => phirong trlnh f(x) = 0 luon co nghiern (dpcm)
Bili 9: Cho phirong trlnh: ax2 + bx + c = 0 (a ;t:. 0)
Chung minh rang neu cac s6 a, b, c thea man dhg tlnrc sau: 2a + 3b + 6c = 0 thl phuong trlnh tren luon co nghiern E (0,1)
Giiii
16
each 1: Sit dung tinh lien tuc:
D<Jt: f(x) = ax2 + bx + e .Ta thay f(x) luon lien tuc Vx E (0,1)
Laic6: f(O)=e'f(~)= 2(2a+3b)+ge= 2(-6e)+ge =-!e
. '3 9 9 3
=> f(O).f(~) = -ie2 s °
=> phuong trtnh f(x) = ° luon c6 nghiem x E ( 0, ~) c (0,1) each 2: Sit dung dinh ly Lagrange:
Xet ham s6 Ftx) = ! ax" + ! bx2 + ex
3 2
Ta tha y F(x) luon lien tuc tren [0, 1] va F' (x) = ax2 + bx + e luon t5n tai tren (0, 1). Theo dinh ly Lagrange thl t5n tai s6 a E (0,1) sao cho:
F(1) - F(O) = F'(e) <:::>!a +! b + e = aa2+ ba + e 32'
2a + 3b + 6e 2 b
<:::> = aa + a + e
6
Theo gia thitt: 2a + 3b + 6e = 0 => aa2 + ba + e = 0
a la nghiern cua phuong trlnh ax2 + bx + e = ° => (dpcm)
Bai 10: Cho m dirong va ba s6 a, b, C E R thea: _a_ + _b_ + ~ = 0. Chung m+2 m-i I m
minh ding phirong trlnh: ax2 + bx + e = ° c6 it nhat mQt nghiern E (0,1)
Giai each 1: Sfr dung dinh ly Lagrange.
Xet ham s6 : F(x) = _a_ xm+2 + _b __ xm+l + ~ x'"
m+2 m i I m
Ta thay F(x) luon lien tuc (0, 1)
F'(x) + axm+l + bx'" + cx""! = xm-1(ax2 + bx + e) lu6n t5n tai tren (0, 1). Theo dinh ly Lagrange thl 3a E (0,1) sao cho:
F(1) - F(O) F'() abe m-l( 2 b )
---- = a <:::> -- + -- + - = a aa + a + e
1-0 m+2 m+1 m
Theo gia thiet: _a_ + _b_ + ~ = ° nen am-1 (aa2 + ba + e) = ° m+2 m e l m.
(aa2 + ba + e) = ° => a langhiern cua phirong trlnh: ax2 + bx + e = ° each 2:
17
Xet f(x) = ax2 + bx + c
+ Ne'u a = 0 ~ f(x) = bx + c
+ Ne'u b :f:. Othl f'(x) = 0 ¢:> x = -~
b
T' .? h'''' c m m (0 1)
iJ' gla t let ~ -- = __ ~ x = __ E ,
b m s- I m+ I
+ Neu b = 0, til' gia thiet e = 0 ~ f(x) = 0 luon dung '\Ix E (0,1)
Til' hai di~u kien tren ta e6 ke't luan: f(x) = 0 luon co nghiern x E (0,1) + Ne'u a :f:. 0: Ta e6: f(O) = 0
f (m + 1) = (m + 1)2 (_a_ + _b_ + c m + 2 )
m + 2 m + 2 m + 2 m + 1 . (m + 1)2
(m + 1)2 (c m + 2 ) c
= m + 2 - m + c. (m + 1)2 = - m(m + 2)
~ f(O).f(m + 1) = _ c2 < 0 va, b,c,m > 0
m + 2 m(m + 2)
phirong trlnh f(x) = 0 luon e6 nghiern x E ( 0, : : ~) c (0,1)
Bai 11: Gid sa a + b + c = 6. Chung minh rhng t6n tai m9t trong ba phirong trinh
x2+ax+1=0 (1)
sau e6 nghiern: x2 + bx+ 1 = 0 (2)
x2 + CX + 1 = 0 ( 3 )
1.
Giai
(1) e6: ~1 = a2 - 4; (2) co: ~2 = b2 ~ 4; (3) co: ~3 = c2 - 4 ~ ~1 + ~2 + ~3 = a2 + b2 + c2 -12 = a2 + (b + C)2 - 2bc -12 ~ T = a2 + (6 - a)2 - 2bc -12
Neu be > 0 ¢:> {h> 0 v {b < 0
c s O . c c O
N'" {b > 0 hi h bA' d~ h' CA • b (b+C)2 (6_a)2
a. eu c > 0 t 1: t eo at ang t uc OS1: c s -2- = -2-
Suy ra: T = a 2 + (6 - a)2 - 2bc - 12 ~ a 2 + (6 - a)2 - 2 ( 6 ; a J - 12 = a2 + _! (6 - a)2 -12 = ~ (a - 2)2 ~ 0
2 2
18
=> ~l + ~2 + ~3 ;::: 0 => it nha't mQt trong ba sCi la sCi am nen c6 mQt trong ba phirong trinh c6 nghiern => (dpcm)
, {b < 0 '
b. Neu thi tu a+ b + c = 6 => a > 6
c<O
Khi a > 6 tht hi~n nhien phirong trlnh (1) luon c6 nghiern (dpcm)
2. Neu be < 0
Xet bi~u thirc: T = a2 + (6 - a)2 - 2bc -12 Xem T la tam thirc b~c hai dCii vdi in a, ta c6: ~T = 36 - 2(24 - 2bc) = 4bc - 8 < Ova be < 0
=> T luon cung da'u vdi M sCi a2 nic la T> 0 va E R => ~l + ~2 + ~3 > 0
Suy ra: C6 mQt trong ba sCi luon duong => c6 mQt trong ba phirong trlnh da cho c6 nghiern.
T6m lai: vdi a + b +c = 6 thl c6 mQt trong ba phirong trlnh da cho luon c6 nghiern => dpcm.
eHUYEN DE 5: PHUONG TRINH B!e HAl CO HAl NGHIEM THOA MAN MOT DIEU KIEN eHO TRUde
1. Cac di~u ki~n eho truoe thuClng gijp
a. Hai nghiern Xl; X2 thea man mot phirong trlnh; mQt ba't phuong trlnh
b. MQt bi~u thtrc clnra hai nghiern Xl; X2 dat gia tri ldn nha't; be nha't.
c. Nghiemarn; nghiern dirong; t '
d. Nghiern thuoc (a, b) cho tnrdc.
e. Nghiern nguyen.
f. Nghiern nay bang k lh nghiem kia.
2. Phmmg phap ~iai Blti todn l
Hai nghiern Xl; X2 thea man mQt phuong trlnh; mQt ba't phuong trinh:
• Budc 1: Tim m d~ phuong trlnh c6hai nghiern Xl; X2
• Budc 2: Bie'n d6i phirong trinh, ba't phirong trlnh chua (x, + X2); XI.X2.
Sau d6 ap dung dinh 19 Viet; gidi tim m.
• Budc 3: Giao cac di8u kien rang buoc cua m. Ke't lu~n.
Blti todn 2
Bi~u thrrc chua hai nghiern Xl; X2 dat gia tri ldn nha't; be nha't. ..
19
• BIMe J:
• BIMc 2:
TIm m d~ phirong trlnh e6 hai nghiern Xl; X2
Bien d6i bi~u thtrc chua (x, + x2); XI.X2. Dung dinh ly Viet dira bi~u thrrc v~ ham f(m)
D~ tim min, max ella f(m) e6 th~ dung mQt trong cac each sau:
Phan tich f(m) = ±(am + pf + A 2 , Dung khao sat ham s6.
Dung di~u kien e6 nghiern ella phuong trtnh b~e 2. Dung ba't ddng thirc Cosi, Bunhiakopxki
Dung tinh bi chan; tinh nha't ella ba't ddng thuc.
• Budc 3: each J:
each 2: each 3: each 4: each 5:
Bai todn 3: Nghiern thuoc (a; b); am duong. each J: Dung cong thirc so sanh nghiern.
Cach 2: Dung khao sat ham s6 va mien gia trio
Bai too" 4: Nghiern nguyen
Dung tinh ehia he't ; s6 chinh phirong.
Bai todn 5: Nghiern nay bang k l~n nghiern kia:
• Budc J: Gid sil' phirong trlnh b~e hai e6 2 nghiern Xl; X2 •
• Budc 2: Nghiern nay bang k l~n nghiern kia nen:
[Xl = kx, [Xl - kx2 = 0
_ <=> _ . <=> (x, -kx2 )(X2 - kx.) = 0
X2 - kx. X2 - kx, - 0 .
• Budc 3: Bien d6i dAng thirc tren v~ t6ng; tich. Sau d6 dung dinh ly Viet.
2. Rai t~p
Bal 1: Cho phuong trlnh b~e hai:
X2 + 2( m + 1)x + 2m2 - m - 3 = 0 (1)
1. Dinh m d~ phirong trlnh e6 hai nghiern Xl; X2 thoa: 2XI - x2 = m + 5 2. Dinh m d~ phuong trlnh e6 hai nghiern sao eho bi~u thirc:
A = x~ + x~ - 2XI X2 da t gia tri nho nha't; ldn nha't.
3. Dinh m d~ phirong trinh e6 hai nghiern Xl; X2 thoa: xr + X~ = 54(xl + X2)
4.
Do h dll h ' h ' hai h'A h' Xl x2 2m - 3 rzm m e p irong tnn co ar ng iern Xl; X2 t oa: - + -- ~ ---
X2 Xl 10
Giiii
(1) e6 2 nghiern Xl; X2
. 2 2
<=> L\ = (m + 1) - (2m - m - 3) ~ 0
<=> _m2 + 3m + 4 ~ 0 <=> -1 ~ m $ 4 (*)
1. Dinh m d~ phirong trlnh c6 hai nghiern Xl; X2 thoa: 2XI - ~2 = m + 5
20
{Xl + X2 = -2(m + 1)
Theo dinh 19 Viet ta du'<JC: 2
XI,X2 = 2m - m - 3 (2)
, {Xl + X2 = -2(m. + 1) {Xl =-: 1 - ;
Ta co: <=>
2XI - x2 = m + 5 x2 = -3 _ 5m
3
The' vao (2) ta du'<Jc:
<=> 13m2 _ m = 0 <=> [m = 03 (thoa man (*»
9 3 m=-
13
VA /. • , •• ~'1>' 0 3
ay cac gia tn cua m can tim a: m = v m =-
.. 13
2. Dinh m d€ phuong trlnh co hai nghiern sao cho bi€u thtrc:
A = x~ + x~ - 2XI x2 da t gia tri nho nha't; 1(1n nha't.
Ta co: A = x~ + x~ - 2xIX2 = (x, + x2)4! - 4xI,X2
= 4(m + 1)2 - 4(2m2 - m - 3) = -4m2 + 12m + 16 Dl}t f(m) = -4m2 + 12m + 16
.. 3
f (m) = -8m + 12; f (rn) = 0 <=> -8m + 12 = 0 <=> m = -
. 2
Bang bien thien
M
-1
3 2
. 4
f'(m)
+
o
f(m)
25
o ------ --------. 0
Tu bang bie'n thien ta diroc:
max A = max f'(m) = 25khim = ~ 2
min A = min f'(m) = 0 khi m = -1 v m = 4
3. Dinh m d€ phirong trlnh co hai nghiern Xl; X2 thoa: x~ + x~ = 54(xI + x2)
Ta co: x~ + x~ = 54 (x, + X2)
21
¢:> (x, + X2)3 - 3xl·X2(Xl + X2) - 54(xl + X2) = 0 ¢:> (x, + X2) [ 4(m + 1)2 - 3(2m2 - m - 3) - 54] = 0 ¢:> -2(m + 1)( -2m2 + 11m - 41) = 0
[m+ 1 = 0
¢:> 2 ¢:> m = -l(nhan)
-2m + 11m - 41 = O(VQ nghiem) . ,
V~y gia tri m dn tlm la m =-1
4 n· h d~ h ' h 6 hai h·A h' Xl x2 2m - 3 . rrm m e p irong tnn c ai ng iern X1;X2 t oa:- + - ~ ---
X2 Xl 10
Ta co: Xl X2 X~ + X~ (x, + X2)2 - 2XlX2 ( 0)
-+- = = Xl'X2:#
X2 Xl Xl X2 Xl·X2
4(m + 1)2 - 2(2m2 - m - 3) 10
= = (rn e -1)
2m2 -m-3 2m-3
Do d6 ba't dAng tlnrc da cho nrong dirong v(1i:.
10 2m- 3 (7 + 2m)(13 - 2m) 0
---~ ¢:> ~
2m-3 10 2m-3
r' 7
m:s;--
¢:> 2
~ < m s 13 2 2
Ke't hop vdi di~u kien (*) ta dtt<;1c % « m s 4
Bal 2: G9i Xl; x21a 2 nghiem cua phtrong trlnh.. X2 + 2mx + 4 = 0 (1) 1. Hay tinhtheo m cac bi€u thirc sau:
a. A = rx; + F; ; b. B = tjx; + ~
2. Dinh m sao cho xt + x~ :s; 32
3. Dinh m de: (::)' +(::)' ~3
4. Tim gia tri ldn nha't, nho nha't ella bi€u thirc:
T = 2(x~ + x~) + 3(xl + x2) + XlX2 3(x~ + x~) + 2(xl + x2) + 4XlX2
Giiii
1. Theo de bai yeu du thl phirong trtnh (1) phdi e6 nghiem duong:
{6.' ~ 0 {m2 - 4 ~ 0 '
¢:> S > 0 ¢:> -m > 0 <=> I? s -2
P>O 4>0
22
{Xl + X2 = -2m Thea dinh lyViet:
Xl.X2 = 4
a. A = Ft +.J;; => A 2 = Xl + x2 + 2JXl x2
=> A 2 = -2m + 2J.1 = 4 - 2m => A = .J4 - 2m b. B = rx; + tjx;
=> B4 = Xl + x2 + 4~ (Ft +.J;;) + 6JXlX2 = -2m + 4W . .J4 - 2m + 6J.1
=> B4 = 12 - 2m + 4J2.J4 - 2m => B = ~12 - 2m. + 4J2.J4 - 2m
2. Dinh m dG: xt + x~ :s 32
, . 4 4 _ (2 2 )2 2 2 2 _ [( )2 2 J2 2 2 2
Ta co. Xl + x2 - Xl + x2 - Xl X2 - Xl + X2 - Xl X2 - Xl X2
= (4m2 - 8)2 - 2.16
=> xt + x~ :s 32 ¢:> (4m2 - 8)2 ~ 32 :s 32 ¢:> (4m2 - 8)2 :s 64 ¢:> Iml :s 2
• Mat khac d6 (1) c6 nghiern thl m2 - 4 ~ 0 ¢:> Iml ~ 2 Do d6 ta chi nhan m = ±2
V~y gia tri cua m c~n tim la m = ±2
3. Dinh m M: (~:r + ( ~: J ,,3
Ta c6: (~)2 + (X2)2 = xt/ ~~ = (4m2 - 8)2 - 32 ~ 3
x2 Xl Xl.X2 16
¢:> (4m2 - 8)2 ~ 80 ¢:> 4m2 ~ 8 + 4/5 ¢:> Iml ~ ~2 + J5 (nhan) V~y gia tri m dn phai tim la: Iml ~ ~2 + /5
Ta c6: T = 2(x~ + x~) + 3(xl + x2) + xlX2
3(x~ + x~) + 2(xl + x2) + 4XlX2
4.
_ 2(xl + X2)2 + 3(xl + x2) + XlX2 = 4m2 - 3m....; 6 - 3(xl + X2)2 + 2(xl + x2) + 4XlX2 6m2 - 2m:- 4
T' = 10m2 + 40m . "T' = 0 ¢:> [m = 0
(60m2 -2m-4)2" m =-4
23
**Bang bien thien
**
m -00 -4 -2 0 2 +00
T' + 0 - +
7 2
T ~/O \2 !/3
-
3 3 4 Dua vao bang bie'n thien ta diroc:
MaxT =!_ khi 10
m = -4 . MinT =! khi m = 2
, 4
Bal 3: Cho phirong trlnh bac hai: X2 + mx + 1 = 0 (1)
l. Dinh m dE phirong trlnh c6 hai nghiem x,;x2 thea man: (:: r + (:: r > 14
2. Dinh m d~ phirong trlnh c6 hai nghiern Xl; X2 thea man: X~ + ~ ;5;; 2
3. Dinh m E Z d~ phuong trlnh c6 hai nghiern Xl; X2 thea bi~u thtrc:
A (x, - X2 )2 d . ,. A
= at gia tn nguyen.
Xl + X2 + 1
Giai
Phuong trinh (1) c6 nghiern ~ ~. ~ 0 ~ Iml ~ 2 (*)
1. Ta c6: (XI)2 + (X2)2 = xt 2+ ~~
X2 Xl XI,X2
[ 2 J2 2 2
(X~ + X~)2 - 2x~ .x~ _ (Xl + x2) - 2X1X2 - 2xI,X2
= 2 2 - 2 2
XI,X2 XI,X2
{Xl + X2 =-m Theo dinh 19 Viet ta dU'<;1c:
XI,X2 = 1
Dod6: (XI)2 +(X2)2 = (m2-2)2-2 =(m2_2)2_2
X2 Xl 1
~ (:J +(~ r > 14 c> (m2 _2)' -2> 14 c> Iml >.J6
D6i chieu vdi di~u kien (*) ta dU'<;1c: Iml. > J6 Vh gia tri m dn tlm la: Iml > J6
24
2. Ta co: x~ + x~ = (x. + X2)3 - 3XIX2(XI + x2) ~ 2 <=> _rn3 + 3m - 2 s 0 <=> rn ~ -2
D6i chieu vdi di~u kien (*) ta du'qc: [: : ;2
V~y cac gia tri ciia m can tim IA: [:: ;2
Tac6: A= (Xl-X2)2 -4XIX2 = rn2 -4 =-rn-1- 3
Xl + x2 + 1 -rn + 1 -rn + 1
Do do d~ A thl (m - 1) IA u'(1c so' cua 3
rn -1 = -1 rn = 0 (loai theo dieu kien (*))
rn -1 = 1 rn = 2 (thoa (*))
3.
<=> <=>
rn-1 = -3 rn =-2
rn-1 =3
rn=4
V~y gia tri m ch tim IA: rn E {-2,2,4}
Bai 4: Cho phirong trlnh bac hai: 2X2 + 2(rn + 2)x + rn2 + 4rn + 3 = 0 (1) I. Dinh m d~ phirong trlnh t 1) c6 hai nghiern Xl; X2 sao cho bi~u thtrc:
A = X31 + X32 -.!. m" - 23 rn dat gia tri nhO nha't.
6 2 . .
2. Dinh m d~ phirong trlnh (I) co hai nghiern Xl; X2 thea man:
xf + x~ = .!.(rn + 1)(rn + 3)
2
3. GQi Xl; X2 IA hai nghiern ciia (1). Chung minh rang:
Ix, + X2 + 3x,x21" (1 + ~J
Giiii
(1) c6 nghiem Xl; X2
<=> fl' =' _rn2 - 4rn - 2 ~ 0 <=> -J2 - 2 s rn ~ -2 + J2
{Xl + X2 = -Im + 2)
"Theo dinh 19 Viet: 1 2
. Xl,X2 = "2 (m + 4rn + 3)
3 3 1 3 23 . 3 ,1 3 23
1. A = Xl + x2 - - rn - - rn = (x, + x2) - 3xl X2 (x, + X2) - - rn - - rn
6 2 6 2
25
{Xl + X2 = -(m + 2)
Theo dinh 19 Viet: 1 2
Xl.X2 = - (m + 4m + 3) 2
A = -(m + 2)3 +~(m+2)(m2 +4m+3)-.!.m3 _ 23
2 6 2
A = m 2 + 6m - 7 ; A' = m 2 + 6m _ 7 = 0 <=> [m = 1 m=-7
Bang bien thien
m -7 AI
-J2-2
J2-2
1
A
Dira vao bang bien thien ta diroc:
min A = A(J2 - 2) = 79 - 43.[3 dat diroc khi m = J2 - 2
3 . .
T ' . 4 4 - [( )2 2 J2 2 2 2
2. aco. Xl +X2 - Xl +X2 - XlX2 - XlX2
= 1 - .!. (m2 + 4m + 3)2 .
2
Do d6: x: + x~ = .!.(m + l)(m + 3) 2
<=> 1- _!_(m2 + 4m + 3)2 = _!_(m + l)(m + 3) = _!_(m2 + 4m + 3)
2 2 2
. [m = -2-J2
=> m2 + 4m + 3 = 1 <=> m2 + 4m + 2 = 0 <=> In (nhan)
m = -2 + ,,2
3.
[m = -2-J2
V~y gia tri ciia m dn Om la: In
m = -2 + ,,2
, 1
Ta c6: Xl + x2 + 3XlX2 = -(3m2 + 10m+ 5) 2
DiU: f'(rn) = .!.(3m2 + 10m + 5)·, r'(m) = 3m + 5
. 2
, 5
f (rn) = 0 => m = -- 3
26
Bang bie'n thien
M -2-J2
5 3
J2-2 '
f'(rn) 0 +
f(m) i+J2~ 5 Y
--
3 Dira vao bang bien thien, suy ra:
5 3 In
--=:;;f(m)=:;;-+,,2 suyra
3. 2
Bai 5: Cho phirong trlnh: ax2 + x + a - 1 = 0 (1)
Dinh a d~ (1) co 2 nghiern thuc Xl =1= x2 thea man [_!_ - _!_[ > 1
Xl X2
Ghii Phuong trlnh (1) co 2 nghiern phan biet: <=> L\ = 1 - 4a(a - 1) > 0
2 1-J2 1+J2
<=> -4a + 4a + 1 > 0 <=> < a < (2)
2 2
<=> -5a2 + 6a > O(a =1= 1)
<=> 0 < a < 1 v 1 < a <.§. thoa man di~u kien(2)
5 "
VA a "a ". k, r 0 . 1 1 6
ay c C gr tn cua a can tim a: < a <. v < a < '5
\
27
Bai 6: Vdi nhirng gia tri nao cua a thl phuong trlnh: X2 + 1 = x e6 hai nghiern a
thuc Xl; X2 kha e nha u thoa man: Ix~ + x~ 1 > 1: a
Giai
Phuong trlnh X2 + 1 = ~ ~ ax2 - X + a = 0, phuong trlnh nay e6 hai nghiern phan a
biet ~ L1 = 1 - 4a 2 > 0 ~ - -~ < a <.!. va a :;; 0
, 2 2
, 1 2 21 1 (2 2 )2 1
Ta co: Xl - x2 > a ~ Xl - X2 > a2
I
~(XI +X2)2(XI-X2)2 > :2 ~(XI -tX2)2[(XI +X2)2_4xl,X2]> :2 ~_!._-4 > 1 ~ 1--5a2 > o~ _J._ < a <J.-(a:;; 0)
a2 F5 ~
Ke't hop vdi dieu kien ta du<;1e gia tri a eh tlm la:
1 1
-- < 1 < 0 v 0 < a <-
2 J5
BM 7: Cho phirong trlnh: xz. - (2 sin a - l)x + 6 sin" a - sin a-I = 0 Tim a d~ phirong trlnh e6 hai nghiern Xl; X2.
Tim gia tri ldn nha't va nho nha't cua bi~u thtrc: A = x~ + x~
Giai
Phuong trlnh (1) e6 hai nghiern Xl; X2 ~ L1 ~ 0
~ (2sina _1)2 - 4(6sin6 a - sin a -1) ~ 0
~-20sin2a+5~0
1. 1 ~--:::;;sma:::;;-
2 2
[_!: + k21t':::;; a s !: + k21t
6 . 6 1t. 1t
~ ~ - - + kn $, a :::;; - + kn
51t 71t 6 6
- + k21t :::;; a :::;; - + k21t
6 6
Khi d6 ta e6: A = x~ + x~ = (x, + X2)2 - 2XIX2 = -Bsirr' a - 2sin a + 3
E>a t t = sin a thl - 1: s t s 1:
, 2 2
Xet ham sCi : f(t) = -8t2 - 2t + 3
. . 1
f (t) = -16t - 2 ; f (t) = 0 ~ t = -- 8
28
Bang bien thien
t
1 2
1 2
1 8
f(t) + 0
25
f(t) 2 ~ -------0
Dua vao bang bien thien ta du'<Je:
max A = maxf(t) = f( -~) = 2: khi t = -~ => sin u = -~ = -sin 13
min A = f (!) = 0 khi t = ! => sin a = ! => a = 2: + mn (m E Z)
2 226
Bai 8: GQi Xl; X2 la hai nghiern cua phirong trlnh:
~ 2 12
12x- - 6mx + m - 4 + - = 0 (1)
m2
Dinh m d~ bi~u thirc A = x~ + x~ dat gia tri Ion nhft; nho nha't.
Giiii
Phuong trlnh (1) c6 nghiern 144
~ -3m2 + 48 - -2 ~ 0 ~ -3m4 + 48m2 -144 ~ 0
m .
~ 2 ~ m s 2J3 v -2J3 ~ m s -2
JXI + x2 = ~
Theo dinh ly Viet: m; 1 1
XI·X2 = 12 -3'+ m2
3 3 3 ( )( ) m 3
Tac6: A=xI +X2=(XI +X2) --3 XI.X2 Xl +X2 =---
2 2m
. 1 3
A = - + --2 > OV'm =f:. 0 2 2m
Bang bien thien
m -2J3
AI
-2
2
+
A
29
Ova vao bang bie'n thien ta diroc:
max A = 3J3 dat taim = 2J3 ., min A = - 3J3 dat tai m = -2J3
4 . . 4
Bili 9: Cho phirong trtnh b~c hai:
( a 2 - 2a + 2) X2 + 2 ( a - 1) x.a - 3a 2 = 0
1. Chung minh ding vdi va > 0 thr (1) c6 hai nghiern Xl; X2 thoa man: 1 1 2
-+-<-
Xl x2 3
2. Dinh a d~ phirong trinh (1) c6 hai nghiem Xl; X2 sao cho:
XI·X2 < laxll-lax21
Giiii Phuong trlnh (l) c6 hai nghiern Xl; X2
<=> !l.' = a2(a _1)2 + 3a2(a2 - 2a + 2)
= a2(a _1)2 + 3a2 [(a _1)2 + 1] 2! OVa
Do d6 (Ijluon c6 nghiern va .
1 Th d~ bai 1 1 2
. eo e iiI - + - < -
Xl x2 3
<=> Xl + x2 < ~ <=> _! < Ovoi Va » 0 => (dpcm)
XI.X2 3 a
2. Theo dinh ly Viet: XI.X2 = 2 -3a 2 ~ 0
a - 2a + 2
+ Neu a = 0 thl Xl = X2 = 0 nen ba't dhg thtrc khong thoa man.
+ Neu a :t: 0 thl Xl; X2 < 0
Do d6: ba't dAng thtic XI·X2 < laxll-lax21 <=> XI·X2 < lal (lxll-lx2i)
[XI.X2 <lal(XI +x2)(ne'UXI >O'X2 <0)
<=> XI.X2 < -lal(XI + x2)(ne'UXI < 0'X2 > 0)
[ -3a2 I 2a(1- a)
- < al --::----
a 2 - 2a + 2 . a 2 - 2a + 2
<=>
-3a2 2a(1- a)
a2 -2a+2 < -Ial· a2 -2a+2'
30
¢:>[-3a2 < 2Ial.a(1- a) (*)
3a2 > 2Ial.a(1- a)
, [2a(a - 1) < -3a
+ Neu a < 0 thl (*) ¢:>
2a(1 -r- a) < -3a
1
¢:> -- < a < 0
2
, [2a(a -1) < 3a 5
+ Neu a > 0 thl ¢:> ~ 0 < a < -
2a(1 - a) < 3a 2
T6 I· . .e •• "', I' 1 0 0 5
mal: gia tn eua a can tim a: -- < a < v < a <.. 2 2
Bai 10 Cho tgn; tgf3 la hai nghiern cua phuong trinh X2 + ax + b = 0 (1)
1. Hay tinh gia tri cua bi~u thtrc:
T = sin2(o. + f3) + a sinto + f3).eos(o. + f3) + bcos2(o. + f3)theo a va b.
2. Cho a = 7; b = 3. Hay tinh gia tri cua bi~u thirc:
Q = 12xI - x21 + 12x2 - xII v(ji Xl; X2 ill nghiem cua (1) Wong tmg v(ji a = 7; b = 3
3. Cho a = 3m - 3; b = -5m +12 (m la tham so). Dinh m d~ phirong trlnh e6 hai nghiern rna nghiern nay bAng blnh phirong nghiern kia.
Giai
1. Tinh T:
Phuong trlnh (1) e6 hai nghiern tgo; tgf3 ¢:> ~ = a2 - 4b ;::: 0
{tgo. + tgf3 = -a Theo dinh 19 Viet:
. tgn.tgf = b .
each 1:
T = sin2(o. + f3) + a sinf« + f3). coste + f3) + b cos2(o. + f3)
= .!. [1- cos 2(0. + f3) + a sin 2(0. + f3) + (1 + cos 2(0. + f3»] 2
= .!. [1 + b + cos 2(0. + f3)(b -1) + a sin 2(0. + f3)] 2
tgu + tgf3 a
DiU: t = tgto; + f3) = =--
. 1- tgo..tgf3 b-1
1 [ 1- t2 2t 1
Suy ra: T = - 1 + b + (b-1)--2 + a'--2
2 1+t 1+t
='!'[1+b+ b~1 2[Cb-l)2-a2+2a2J]=b
2 (b -1) + a .
31
each 2:
Ta c6: T = sin2(a + 13) + a sin(a + 13). cos(a + 13) + b cos2(a + 13)
=> T = [1 + tg2(a + P)] = sin2(a + P) + a sin(a + 13)· cos(a + P) + bcos2(a + P) cos2(a + P)
= tg2(a + 13) + atg(a + 13) + b
=> T = tg2(a + 13) + atg(a + 13) + b
1 + tg(« + 13)
V ,. t ( A) tga + tgl3 a
01 g a + t-' = = -- .
1 - tg(a + 13) b - 1
a2 a
---::-2 + a.-- + b
The' VaG T ta du'<;1c: T = (b - 1) ~ - 1 = b
1+ a
(b _1)2
2. Khi a = 7; b = 3 ta diroc: X2 -+- 7x + 3 = 0 Vi A = 12xl - x21 + 12x2 - xli ~ Onen
A2 = (2Xl - X2)2 + (2X2 - Xl)2 + 21(2xl - x2)(2x2 - xl)1 = 5(x2 + Xl)2 - 18xl·X2 + 219xl .X2 - 2(x2 + Xl )21
Vi Xl; X2 la hai nghiern cua (2) nen thea dinh ly Viet ta diroc:
{Xl + x2 = -7 => A2 = 5.49 _ 54 ~ 2127 - 981 = 333 => A = ..,1333
Xl.X2 = 3 .
3. Khi a = 3m - 3; b = -Sm + 12 thl (1) trd thanh:
X2 + (3m -- 3)x - 5m + 12 = 0 (3)
Phuong trlnh (3) c6 hai nghiern Xl; X2
2· -1-..)352 -1+..)352
<=> Ll ~ 0 <=> 9m + 2m - 39 ~ 0 <=> m ::; v m ~ ----
9 9
. {Xl + x2 = 3 - 3m
Theo dinh ly Viet ta du'<;1c:
Xl.X2 = 12 - 5m
Khong ma't tinh t6ng qua t, ta c6 th6 coi: x2 = xf
{xr = 12 - 5m Khi d6:
. Xl + x~ = a - 3m
{3xr = 36 - 15m 3 2
<=> 2 => 3xl - 5xl - 5xl = 21
5xl + 5xl = 15 -15m .
<=> Xl = 3 => 12 - 5m = 27 => m = -3 (thoa man di~u kien)
32
Thl't lai:
Khi m = -3 lhl (3) trd tha nh: X2 -12x + 27 = 0
[Xl = 3~, 2
<=> ta thay X2 = Xl
X2 = 9
Vh gia tri m dn tim Iii: m = 3
Bai 11: Cho phuong trtnh: ax2 - bx + b = 0 vdi a, b > 0, co cac nghiern Hi
Xl; X2 ' Chung minh r~ng t6n tai cac s6 aj = ±1; (i = 1,2) sao eho:
&1 + al & + a2 ~ = 0
V~2 V~ ~a
'Bai 12: X6t cac s6 thirc: a, b, e sao eho phuong trinh: ax2 + bx + e = 0 co hai nghiem huf [()I)H- inh " 'I' hA'? h'':? hti T (a-b)(2a-b)
t uoc , ' ay un gra til (in n at cua IOU l lie: =
a(a - b + c)
Giiii
Vdi de s6 thirc a, b, clam eho phirong trlnh ax2 + bx + c = 0 co hai nghiern E [0,1], Goi hai nghiern do la Xl; X2 ' thea dinh 19 Viet ta du<;1e:
Ta co:
(a - b)(2a - b) ( b)( b)
T = (a - b)(2a - b) = a2 = ...:...1_-_;--,;--,,--2_-_;..:-.
a(a - b + c) a - b + c b c
1--+a a
a
33
(1 + Xl + x2 )(2 + Xl + x2) 1 + Xl + X2 + Xl .X2
= 2(1 + Xl + X2 + Xl .X2) + Xl + X2 + X~ + X~ = 2 + Xl + X2 + X~ + x~
1 + Xl + x2 + Xl .X2 1 + Xl + x2 + XI.X2
Khong ma't tinh t5ng quat co th~ gia si't 0 s Xl :os; x2 :os; 1
=> x~ s XI.X2 => 1 + Xl + X2 + XI.~ ~ 1 + x~ + XI.X2
2 2 2 2 2 2
Xl + X2 + Xl + X2 Xl + X2 + Xl + X2 Xl + X2 + Xl + X2 = 1
=> :os; 2 :os; .....=..--:::2"----=--~
1 + Xl + x2 + XI·X2 1 + Xl + Xl + x2 1 + x2 + Xl + x2
=> T s 2 + 1 = 3 => Tmax = 3
D"" " " ? 1
au = xa y ra ~ Xl = X2 =
Bai 13: Cho phuong trlnh: X2 + ax + b - 2 = 0 v6'i a,b E R
Gia si't phuong trtnh co nghiem kep a ~ 2 . Tim gia tri nho nha't cua A = a 2 + b2
Giai Phuong trtnh co nghiern kep a ~ 2
{_~. > 2 {a :os; -4 {a :os; -4
~ 2- ~ ~ 1
~ = 0 a2 - 4(b - 2) =0 b = "4a2 + 2
(1 )2 1
A = a2 + b2 = a2 + _a2 + 2 = _a4 + 2a2 + 4
4 16
A'=~a3+4a=a(~a2+4); A=O~a=O Bang bien thien
Tit d6:
a -00
-4
A
+ 00 ________
52
AI
Dl,l'a vao bang bie'n thien ta duoc: min A = 52 dat tai {~ : ~4
Bai 14: Cho phu'dng trtnh: X2 - (3 sin a - cos a)x - 4 - 4 cos 2a = 0 (1)
Dinh a d~ phirong trmh e6 nghiern sao eho t5ng blnh phirong hai nghiern cua phu'dng trlnh dat gia tri Ion nha't va nho nha't.
Phuong trlnh (1) e6 nghiern .
34
<=> ~ = (3 sin a - cos a)2 + 4 (4 - 4 cos 2a) ;:?: 0
9 1- cos2a .
<=> - (1 - cos 2a) + - 3 sin 2a + 16 + 4 cos 2a ;:?: 0
2 2
<=> 21- 3 sin 2a :::: 0 luon Iuon dung va E R Do d6 (1) luon c6 hai nghiern phan biet.
G9i Xl; X2 11'1 hai nghiern cua phirong trmh, ta c6:
A = x~ + x~ = (Xl + X2)2 - 2xI,X2 = (3 sin a - cos a)2 + 2 (4 + cos 2a)
= ~ (1 - cos 2a) + .! (1 + cos 2a) - 3 sin 2a + 8 + 2 cos 2a
2 2
= -2cos2a - 3sin2a + 13
Theo btt d~ng thirc Bunhiakopxki:
1-2 cos 2a - 3 sin 2al s ~=[ (-_-2-)2-+-(---3-)2-=J-( s-i-n-2 -2a + cos'' 2a) = .J13
=> -.J13 ~ -2 cos 2a - 3 sin 2a s .J13
=> 13 - m s 13 - 2cos2a - 3sin2a ~ 13 + Ji3 => Amax = 13 + .J13 ; Amin = 13 .: Ji3
Bal 15· Cho phirong trlnh: X2 + 2 (a - 3) X + a -13 = 0 (1)
I TI~ a :::: 1 d~ nghiern ldn nha't cua phuong trlnh nhan gia tri nho nh(t.
Giai
. ( 7)2 39 .
Taco: ~ = a - "2 + 4 > 0 va E R
=> (1) luon eo hai nghiern phan biet: Xl; X2 G9i nghiern ldn nh(t eua phirong trlnh la: x2 =3-a+.Ja2 -7a+22
. 2a-7 2a-7-2.Ja2-7a+22
x2 = -1 + = ---,=:;=======--
2.Ja2 -7a + 22 2.Ja2 -7a + 22
x~ = 0 <=> 2.Ja2 -7a + 22 = 2a-7
{2a -7:::: 0
<=> 4 ( a 2 - 7 a + 22) = 4a 2 - 28a + 49
{ 7
>- .
<=> a - 2 vo nghiern => x2 < OVa z 1.
88 = 49
35
Bang biC'n thien
x
6
o
Xl 2
X2 ________
Dua V~lO bang bien thien (X2 )min khong t6n tai a. V~y khong eo ghi tri a ~ 1 thea d6.
• Chu y: V(1i bai toan tren thl nghiern Ion nha't nhan gia tri ldn nhat ~ a = 1 khi do amax = 6
V(1i diu hoi nhu v~y ta eo the lam each khac nhtr sau:
GQi Xo la nghiern ldn nha't cua phirong trlnh thi:
x~ + 2(a - 3)xo + a - 13 = 0 ~ (2xo + 1)a = 13 + 6xo - x~
13 + 6xo - x~ ( 1 1 khA l' hie ? (1»
=> a = v Xo = -- ong a ng tern eua
(2xo + 1) 2
,,., 1 13 + 6xo - x~ 1 12 + 4xo - x~ 0
v i a ~ => ~ ~ ~
(2xo + 1) 2xo + 1
~lxo1~-2 =>(xo) =6
--<xo~6 max
2
Khi (Xo )max = 6 the' vao phirong trlnh ta ducc: a = 1
Bai 16: TIm ta't cit cac gia tri cila m sao eho phuong trlnh:
X2 - (2m + 1)x + m2 + m - 2 = 0
eo hai nghiern X,I; x2 thea: 0 < Xl < 2 hay 2 < x2 < 5
Giai
Ta eo : ~ = 9 > 0 "d m => phirong trlnh luon eo hai nghiern phan biet.
[Xl = m-1
. R5 rang Xl < x2"dm x2 = m + 2
Yeu du bai toan dmrc thoa man:
{o < m - 1 < 2 {1 < m < 3 1 3
~ ~ ~ <m<
2<m+2<5 O<m<3
Vay cac gia tri cua m cdn tlrn la: 1<m<3.
Bai 17: Cho phirong trlnh: X2 - 219 - log, m] X + 319 - log, m] = 0
Dinh cac gia tri de m9t trong cac n~hit;m cua phirong trtnh l(1n hon 3 con nghiern kia nho hon 3.
36
Giiii
Yeu e~u bai toan du<)e thea man <=> Xl < 3 < X2 <=> f(3) < 0
<=> 9 - 619 -log3 ml-t 319 -log3 m] <0 <=> 19 - log, ml > 3
<=> [9 - log, m < -3 <=> [lOg3 m > 12 <=> [m > 312
9 - log, m > 3 log, m < 6 m > 36
Ke't hop vdi di~u kien m > 0 => [0 < m1::f la cac gia tri m dn tlm. m c- S
Bal 18: Cho hai phirong trlnh: X2 + 3x + 2a = 0 va X2 + 6x + 5a = 0 J
Dinh a de hai phirong trtnh co nghiern xen ke nhau.
Giiii
Dat f(x) = X2 + 3x + 2a; f(x) = 0 co hai nghiern Xl; X2
g(x) = X2 + 6x + 5a ; g(x) = 0 co hai nghiern x3; X4
[Xl < X3 < X2 < X4 Hai phirong trlnh co nghiern xen ke nhau <=>
X3 < Xl < x, < x2
<=> f'(x, Lf'(x,) ~ 0 <=> (x~ + 3X3 + 2a) (x~ + 3x4 + 2a) < 0
<=>(Xs.X4)2 +3x3.x4 (Xs +X4) +2a(~ +x~) +9x3.X4 +6a(x3 +X4) +4a2 < 0 <=> 9a 2 - 9a < 0 <=> 0 < a < 1
Vay cac gia tri ctia a dn tlm la: 0 < a < 1 .
Bai 19: Tim a, b E Z sao cho phuong trlnh: X2 + ax + b = 0 co hai nghiern tho a man: -2 < Xl < -1 va 1 < x2 < 2
Giiii
Theo di~u kien d~ bai:
{-2 < x, < -1 => {-1 < Xl. + x2 < 1 1 < x2 < 2 -4 < Xl .X2 < -1
{Xl + X2 = -a {-1 < -a < 1 {-1 < a < 1
Theo dinh 19 Viet: b => b <=> b 1
X1.X2 = -4 < < -1 -4 < <-
Vi a, b E Z => a = 0 va b = {-3, -2}
{a = 0
+ Ne'u => (1) co nghiern X = ±.J3 thoa man d~ bai.
b = -3 .
37
+ Neu {~ : ~2 =:> (1) c6 nghiern x = ±J2 thoa man d~ bai.
VA a . , .? b ~ , r {a = 0 {a = 0
ay c c gia tn cua a, can tim a: b = -3 v b = -2
Bai 20: Dinh m d~ phirong trlnh sau c6 4 nghiern phan biet:
1
- x2 -Ixl + m = 0 (1) 4
Giai
cu« 1:
1 ' Phuong trlnh (1) <=:> m =Ixl- - x2 4
{ 1 2"" 0
-x--x neux<
1 -
Ve Parabol (P): y = Ixl- - x2 = 4
4 1 2"" 0
x--xneux ~
4
y
1
Dua vao dd thi suy ra: phirong trlnh da cho c6 4 nghiem phan biet khi va chi
khi dirong thhg y = m cilt (P): y = Ixl- ~ x2 tai 4 di~m phan biet <=:> 0 < m < 1
Vh cac gia tri ciia m dntim la: 0 < m < 1.
Cach 2: Dung cong thtrc so sanh nghiern:
D~t t = Ixl ~ 0
(1) trd thanh f(t) = .!.t2 - t + m = 0 (2) 4
D~ (1) c6 4 nghiern phan biet thl (2) c6 hai nghiem phan biet dirong: '
<=:> {!: ~ <=:> {~-~> 0 <=:> 0 < m < 1
8>0 1<0
38
·2·
Bai 21: Cho phirong trlnh: (rn + 1) (fx + 1) - 2m (fx + 1) + 2m = 0 (1)
Dinh m d~ phirong trlnh co mot nghiern thuoc [0,4]
Giiii
Di). t t = J";. + 1 ; vdi 0 ~ x ~ 4 => 1 ~ t ~ 3
Phuong trlnh (1) trC1 thanh: (m + 1)t2 - 2mt + 2m = 0 (2)
• (fng voi mot nghiern t E [1,3] thl phuong trlnh t=J";. + 1 co dung mot nghiern x E [0,4] nen:
+ Phuong trlnh (1) co mot nghiern x E [0,4] thl (2) co mot nghiern t E [1,3]
{fi' = 0 ![ : : ~2
<=> m <=> <=> m = -2
to = - E [1,3]. m 3
m i I 1~--~
m s- I
Vh m = -2 thea man d~ bai.
+ Phuong trinh (2) co nghiern t1; t2 thoa man:
[tl < 1 s t2 < 3 1 < tl s 3 < t2
• Ne'u (2) co nghiern tl = 1 thl tit (2) m + 1- 2m + 2m = 0 => m = -1, gia tri nay khong thea man (vl m =I:- 1)
• Ne'u (2) co nghiem t2 = 3 thl.tir (2) => 5m + 9 = 0 => m =-~
,. 9 I ? tha h 4 2 18 t 18 0
VOlm=--th(2)trd an --t +- --=
5 5 5 5
<=> _2t2 + 9t - 9 = 0 <=> [t2 :: (loai VI tl E [1,3] tl -"2
• 'Con lai phuong trlnh (2) co hai nghiern t1; t2 thoa man:
[tl < 1 s t2 < 3 <=> f(1).f(3) < 0 1 < tl ~ 3 < t2
9
<=> (rn + 1)(5m + 9) < 0 <=> -- < m < -1 5
Yay cac gia tri cua m c~n tim la: -~ -cm < -1 v m = -2
.. 5
39
Bai 22: Cho phirong trlnh: sin 3x - m cos 2x - (m + 1) sin x + m = 0 (1)
Dinh m d~ phuong trlnh co dung lam nghiern E (0, 31t)
Giiii
Phuong trinh (l) <=> sin x (4 sin2 x - 2m sin x + m- 2) = 0
{sin x = 0
<=> 4 sin 2 x - 2m sin x + m - 2 = 0 (2)
• V oi sin x = 0 <=> x = k1t (k E Z)
Khi 0 < x < 31t => 0 < kn < 31t <=> 0 < k < 3=> k = [1,2] =>. nghiern x = {1t, 21t}
Dat: t = sinx;-1 s t $1
Phuong trlnh (2) tn1 thanh: 4t2 - 2mt + m - 2 = 0 (3)
+ U ng vdi mot nghiern tl E (0,1) thl phirong trlnh 1 = sinx co dung bon nghiern x E (0, 31t)
. ~
+ a ng vdi m<;>t nghiern t2 E (-1,0) lhl phirong trlnh t = sinx co du ng hai nghiern
X.E (O,31t) nen M (1) co dung tam nghiern x E (O,3n) thl (2) co dung sau nghiern x E (o,31t) va khac cac nghicm s6 1t va 21t. Di~u nay xay ra khi va chi khi phirong trinh (3) co nghiern t1; t2 thoa man: -1 < tl < 0 < t2 < 1; hoac nghiern tl E (0,1) con t2.= 1.
a. NC'u phirong trtnh co nghiern t2 = 1 thl ttr (3) => 4 - 2m + m - 2 = 0
[t = 0 ~ (0 1)
=> m = 2khim = 2 thl (3) tro thanh: 4e -4t = 0 => 1 ,
t2 = 1
b. Phuong trmh (3) c6 hai nghiern t1; t2 thea: -1 < t 1 < 0 < t2 < 1
{f(O) < 0 {m - 2 < 0
<=> f(1) > 0 <=> 2 - m :::. 0 <=> - ~ < m < 2 f(-1) > 03m + 2> 0
V~y gia tri ciia m dn tim la: -~ < m < 2
Bai 23: Dinh a d6 phirong trlnh sau co nghiem nguyen:
x2 - ( 3 + 2a)x + 40 - a = 0
(1)
Gi8i"
Phuong trinh (1) co: L\ = (3 + 2a)2 - 4 (40 - a) = 4a2 + 16a -151
40
Dc (I) co nghiern nguyen thl dieu kien dn Hi: ~ phai lit chinh phirong. => A = n2 => 4a2 + 16a -151 = n2 (vdi n E Z)
<=:> (2a + 4)2 - n2 = 167 <=:> (2a + 4 - n)(2a + 4 + n) = 167 (*)
VI 167la s() nguyen to' nen d~ng thric (*) xa y ra
{2a + 4 - n = 1 h {2a + 4 - n = -167 oac
2a+4+n=167 . 2a+4+n=-1
<=:>
{2a + 4 - n = -1 {a = -44
2a + 4 + n = -167 <=:> n = 83
[a = 40 . , .• I' 40 44
<=:> => gia tn eua a a: a = . v a = -
a = -43,5(lo~i Yin ~ Z) .
Kicm tra:
+ Khi a = 40 thl (1) <=:> X2 - 83x = 0 <=:> [x = 0 (nhan) x = 83
+ Khi a = _ 44 <=:> [x = -1 (nhan) x = -84
Ketlu~n: Gia tri ciia a dn tim la: a = 40; a = :-44
Bai 24: Cho phirong trlnh: mx2 - mqx + q = 0 (m =t- 0)
1. Dinh hi; thirc lien h~ giua m va q d~ hai nghicrn cua plnrong trinh trcn la: sin va tang ella cung mot goc a .
2. Tinh m then a .
3. Cho mq = _.!. Tinh a . 2
Giai
l. GQi hai nghicrn ella phirong trtnhla XI;X2.
Theo dinh ly Viet ta diroc:
{Xl + x2 = sin a + tga = q {Sin a + tga = q
. q <=:> _ sin a + tga _ 1 + cos a
Xl·X2 = SIn a.tga = - m - . t - .
m sma ga . sma
a a
2tg -2- 2tg --
_----'=-- + 2 _ q ( * )
a 'U
1 + tg2__ 1 - tg2 ---
<=:> 2 2 The' van (*)1a diroc:
1 a 1
m=--=>tg-=-
tg~ 2 m
2
41
2 1
m 1 + 2 m 1 = q <=> 4m3 = qtm" -1) 1+- 1--
m2 m2
(Day Iii M thirc lien M gitra q vii m)
2. Theo tren ta diroc: m = _1_ = cot g a
. t a 2
g-
2
3. Khi mq = _.! => q = _ _l__ = _ 1
2 . 2m 2cotg a
2
Theo dAng thirc d ea u (1)
4m3 = qim" -1) <=> 4cotg3 a = _ 1 (cotg4 a -1)
2 2cotg a 2
2
4 a a 1 21t
<=> 9 cot g - = 1 <=> cot g - = ± 10 => a = ± - + k21t
2 2 ",3 3
Vh g6c a dn tim 18.: a = ± 231t + k21t (k E Z)
Bai 25: GQi Xo lit nghiern ciia phirong trlnh: ax2 + bx + c = 0 (a *- 0)
a. Chung minh r~ng: Ixol s 1 + max {I:I;I;-I}
, 21acl + b2
b. Chung minh rang: Ixol ~ labl (a.b *- 0)
Giili
a. Chung minh rang: Ixol s 1 + max{I;I;I;I}
Ta e6: 1 + max{I:I;I;-I} ~ 1 nen:
+ Ne'u IXol s 1 thi ba't dAng tlnrc (1) luon dung. + xer Ixo I > 1 :
• Xo 18. nghiern cua phirong trtnh: ax2 + bx + c = 0 (a *- 0)
2 2 b c
=> -axo = bx, + c <=> - Xo = - Xo +-
a a
42
X2 -1 X2
Mat khac: M(lxol+1)=M'1 01 <Milo
Xo -1 Xo -1
Suy ra: Ixol2 s M 1 I~ ~ 1 s 1 r ~ IXol s M + 1
Xo -1 Xo -1
Hay Ixol s 1 + max{I:I;I~I} ~ (dpcm)
Chung minh rang: Ixol s 2Ial:~ b2 = 21~1 + 1:1
b.
G·• • A hai h' A • h ' h la -b - ~b2 - 4ac ia sit mot trong ar ng rem cua p irong trtn a: Xo = ------
2a
~Ixol = .s., ",b2 -4ac ~.!.Ibl+ Jb2 -4ac
2a 2a 2 a 4a2
11bl ~2 c
~Ixol~--+ --~
2 a 4a2 a
C~ hii . h 11bl ~2 c· 21cllbl
an c ung mm : - - + - - - ~ - + -
2 a 4a2 a b a
Th~t v~y, ba't dAng thtrc (*)
~ ~~1:1+21~1
¢:> ~ - ~ s ~ + 21~1 + 4 £ ¢:> - ~ ~ 21~1 + 4 £
4a2 a 4a2 a b2 a a b2
21acl + b2
Ro rang ba't dAng thrrc nay ludn dung.Vgy: Ixol ~ labl (dpcm)
(*)
Bai 26: Cho phuong trlnh b~c hai: ax2 + bx + c = 0 (a "* 0). Chung minh rAng di~u kien: (k + 1)2 ac - kb2 = 0 (k "* 0) la di~u kien ch va dii d€ phirong trlnh co hai nghiern, trong do nghiern nay bang k Ih nghiem kia.
Giai
Gia sii' phirong trlnh tren co nghiern; nghiem nay bhng k IAn nghiern kia:
[Xl = kx2 [Xl - kx2 = 0 ( . ) ( )
_ kx ¢:> _ kx _ ¢:> Xl - kx2 . X2 - kx, = 0
X2 - I X2 I - 0
43
<=:> Xl ,X2 (1 + k2) - k (xi + xn = 0
~ X1.X2 (1 + k2) - k (Xl + X2)2 + 2kxl,X2 = 0
{Xl + X2 =-~
Thco dinh 19 Viet: _ c a
XI,X2 --
a
XI.X2 (1 + k2) - k (Xl + X2)2 + 2kxl,X2 = 0
c ( 2 \ (b)2 c
<=:> ~ 1 + k ) - k - + 2k. - = 0
a a a
¢:> ac(1 + k2) - kb2 + 2kac = 0 ¢:> ac(k + 1)2 - kb2 = 0
Dao lai: Khi ac(k + 1)2 - kb2 = 0 => ac = kb2 2
(k + 1)
kb2 Ta co: ~ = b2 - 4ac = b2 - 4--'---
(k + 1)2
[(k+ 1)2 - 4kJ b2 I
= b2 :=' (k + 1)2 > 0 Vk::;:. -1
(k+1)2 (k+1( -
Suy fa phuong trinh ax2 + bx + c = 0 co hai nghiern x.: X2
CHUYENDE 6:
QUAN Ht NGHltM CVA HAlpHUONG TRiNH B~C HAl
1. Cac bili toan
a. Bili toan 1: Cho hai phuong trlnh h~c hai: al X2 + b, X + ci = 0
a2x2 + b2x + c2 = 0
Tim cac gia tri ella tham s6 m 00 (1) va (2) co nghicm chung. Pill/dug pluip:
• Budc I: Goi Xo Ia nghicrn chung cua hai phirong trlnh thl:
{aIX~ + b.x, + ci = 0 a2xO + b2x{) + c2 = 0
• Butte 2: Gia i h9 trcn tim xO' Suy ra m.
• Budc 3: The' gia tri ciia m vao (I); (2) 06 ki0m tea.
44
(1) (2)
b. Bai toan 2:
Tim cac gia tri ella tham s6 m d~ (1) va (2) e6 nghiern sao eho mQt nghiern . ella phirong trlnh (1) bang k I~n mot nghiern ella phuong trtnh (2)
Phuong phdp
• Birdc I: G9i Xo Ia mQt nghiern ella (2) thl kxo(k :t: 0) Ia mot nghiern cua (1). Khi d6, Xo la nghiern ella M.
{a2X~ + b2xO + c2 = 0
a, (kxo)2 + bl (kxo) + cI = 0
• Budc 2: Gia i M tren tim xo. Suy ra m .
• Birdc 3: UlY gia tri ciia m the' vao (1) va (2) d6 ki6m tra.
c. Bai toan 3
Tim cac gia tri cua tham s6 m d~ (I) va (2) urong dirong Phuong phdp:
Sit dung dinh nghia VI hai phuong trlnh nrong duong.
2. Bai U)p
Bai 1: Cho hai phirong trlnh: x2 + x - m = 0 (1)
x2 - mx + 1 = 0 (2)
Dinh m d~ phirong trlnh (1) va (2) e6 nghiern chung.
Giai
GQi Xo Ia nghiern chung cua hai phuong trlnh thl Xo thea hi;:
{X2 + ~ - m = 0 (*) ..
~ 0 _ ::)(l+m)xo-(m+l)=O (3)
xo-mxo+1-0
+ Ne'u m = -1 thl (3) nghiern dung vdi VXo
nhirng khi m = -1 thtphuong trlnh (1): x2 + x + 1 = 0 va plurong trlnh (2): x2 + x + 1 = 0
Hai phuong trinh nay VQ nghiern nen m = -1 khong thoa man. + Ne'u m :t: 1 thl (3) e6 nghiern duy nha't Xo = 1 .
Khi Xo = 1 the' vao (*) ta du'<;1e: m = 2
V oi m = 2 thl tit (I): x2 + x _ 2 = 0 ¢:> [Xl = 1 x2 =-2
va (2): x2 - 2x + 1 = 0 ¢:> xa = 1 Ta tha'y Xl = xa = 1 nen m = 2 thea ~e.
V~y gia tri cua m e~n tim la: m = 2
45
Bai 2: Cho hai phirong trlnh:
x2 - ( 2m + 1) x + 2m + 2 = 0 (1)
va x2 - 2mx + m + 1 = 0 (2)
Dinh m d~ hai phirong trinh co nghiern chung.
Giai
GQi Xo la nghiern chung cua hai phuong trlnh thl Xo thO a M:
{x~ - (2m + 1) Xo + 2m + 2 = 0
=> Xo == m + 1 x~ - 2mxo + m + 1 = 0
Khi Xo = m + 1 the' vao (3) ta diroc:
(m + 1)2 - 2m ( m + 1) + m + 1 == 0
~ (m + 1)(m - 2) = 0 ~ m = -1 v m = 2
+ Vdi m = -1 ta diroc: (1) ~ x2 -r x = 0 ~ X =. -1 v x = 0 (2) ~ x2 + 2x = 0 ~ x =-2 v x = 0 => nghiern chung x == 0 => m = -1 thea de.
+ V oi m = 2 ta diroc: (1) <=> x2 - 5x + 6 = 0 ~ x = 2 v x = 3 (2) ~ x2 - 4x + 3 = 0 ~ x = 1 v x = 3 => nghiern chung x = 3 => m = 2 thea de.
V~y phtrong trlnh (1) va (2) co nghiern chung ~ m = -1 v m = 2
va
x2 + 2 (m + 1) x + 3m + 1 = 0
(2)
Ba] 3: Cho hai phuong trmh:
x2 + (3m + 1) x + 2m + 2 = 0 (1)
Dinh m d~ hai phirong trlnh tren co nghiern chung.
Giai
GQi a la nghiern chung cua hai phirong trinh thl a phai thea M:
{a2+(3m+1)a+2m+2=O {(m-1)a=m-1 (3) a 2 + 2 (m + 1) a + 3m + 1 = 0 ~ a2 + 2 (m - 1) a + 3m + 1 = 0 (4) + Ne'u m = 1 thl (3) vo s6 nghiern: tit (4) =>a = -2
+ Ne'u m -:;:. 1 thl (3) co nghiern duy nhft: a = 1 the' vao (4) ta du<jc:
5m + 4 = 0 ~ m = _ 4
5
T ' 1· C' . , .? ", 1)1 1 4 om at: ac gia tn cua m can tim a: m = v m = --
.. 5
46
Bai 4: Cho hai phirong trlnh: x2 + ax + 1 = 0 va x2 + bx + 2 = 0
1. Tim di~u kien cua a, b d~ hai phirong trinh c6 nghiern chung.
2. Vdi di~u kien tren, tim gia tri nho nha't cua A = lal + Ibl
Giai
1. GQi Xo 1a nghiern chung cua hai phtrong trlnh tht Xo thoa h¢:
{X022 + axo + 1 = 0 (1)
~(a-b)xo=l (*)
Xo + bx, + 2 = 0 (2)
+ Ne'u a = b thl phtrong trtnh (*) va nghiern nen (1) va (2) khong c6 nghiem chung.
+ Ne'u a ;t. b thl (*) ~ Xo = _1_ the' vao (1) ta duoc:
b-a
1 2 + _a_ + 1 = 0 ~ 2a2 + b2 - 3ab + 1 = 0 (**)
(a-b) a-b
Vdi di~u kien (**) ta ki~m tra lai hai phirong trlnh (1) va (2) luon c6 nghiern chung xO'
2. Tim min A = lal + Ibl
Tltdhg thirc (**) ~ 3ab = 2a2 + b2 + 1 ~ ab > 0 ~ lal.lbl = ab Lai c6: A = lal + Ibl ~ Ibl = A -Ial
Tu (**) ~ 21al2 - 3lal.lbl + b2 + 1 = 0
¢:> 21al2 - 31al (A -Ial) ... (A -lal)2 + 1 = 0 ¢:> 61al2 - 5Alal + A2 + 1 = 0
D~t Y = lal;Y > 0
Thl: 6y2 - 5A.y + A2 + 1 = 0
Dinh A d~ phirong trlnh nay c6 nghiem urong dirong
¢:> {! ~ ~ ¢:> {A 2 ~ 24 ~ A ~ 2J'B
8>0 A>O
Da' u "=" xii y ra ¢:> 6y2 - lOJ'By + 25 = 0 5J'B
a=--
6
- ~
5../6
a=---
6
5J'B
¢:>y=--~
6
b = 7J'B 6
b = _ 7J'B 6
47
Rai 5: Cho hai phirong trlnh: x( x - 1) = m + 1
va X4 + (x + 1)2 = m2
Dinh m d6 hai phirong trtnh c6 nghiern chung.
(1) (2)
Giai Phuong trlnh (1) <=> x2 - (x + 1) = m
{a = x2 > 0 2
I)~ t: - => a = (b - 1 )
b = x + 1
(*)
Khi d6 (1) va (2) tru thanh: a - b = m
(3)
{a - b = m {a - b '"' m {a - b = m
<=> a + b = ±m <=> a + b m v a + b = -m
1.
{a - b = m {a = m .
V oi <=> th0 v iw( *) ta duoc: m = 1
a r b e m b=O
{a = 1 {X2 " 1
Khi m = 1 thl: <=> <=> x = -1 (nghiern chung)
b = 0 x = --1
2.
=>m = 1 tho a d~.
{a - b = m {a = 0
Vdi . <=>
a+ b = -m b =-m
2 fa = 0 {x = 0
[m-i I] =o<=>m=-l=>l <=> <=>x=O
b=l x r= O
=> m =- 1 thou d~·.
. [m = 1
V~y phlfdng truth (I) va (2) co nghicrn chung <=> m = -1
48
Bai 6: Chung minh rAng: ne'u hai phirong trlnh:
x2 + ax + b = 0 va x2 + ex + d = 0
C6 nghiern chung thl: (b - d)2 + (a - e)(ad - be) = 0 (vdi a -:f:: c)
Giiii
GQi a Ia nghiern chung cua hai phirong trlnh thl a thoa he:
{a2 + ac + b = 0 (1) ( ) b d
=> a-e a+ - =0
a2 + co + d = 0
d-b
<=> a = -- (a -:f:: c) a-e
d - b . (d - b)2 d - b
The' a = -- vao (I) ta ducc: -- + a.-- + b = 0
a-e a-e a-e
<=> (b - d)2 + a (d - b)(a - c) + b (a - e)2 = 0 <=> (b - d)2 + (a - e)(ad- ba + ab - be) = 0 <=> (b - d)2 + (a - e)(ad - be) = o (dpcm)
Bai 7: Cho hai phirong trlnh: x2 - 2mx + 4m = 0
(1)
va x2 - mx + 10m = 0 (2)
Dinh m de phirong trlnh (1) va (2) c6 nghiern; trong do m(;t nghiern cua phirong trlnh (2) bang hai l~n met nghiern cua phirong trlnh (1).
Giiii
'2 [m::;; 0
Phuong trlnh (1) co nghiern <=> ~l = m - 4m ~ 0 <=>
. m z d
'2 [m::;; 0
Phuong trlnh (2) co nghiern <=> ~2 = m - 40m ~ 0 <=>
m~40
=> gia tri chung: [m s 0
m~40
GQi a la m(;l nghiern cua phirong trlnh (1) thl (2a) la m9t nghiern ella phircng trlnh (2). Do do, ta co he:
{a2-2ma+4m=o {a2-2ma+4m=o
(2a)2 - m(2a) + 10m = o<=> 4a2 - 2ma + 10m = 0
- (1)
(*)
(2)
49
+ Neu m = 0 thl (3) nghiern dung con (4) c6 nghiem (l = 0; thoa di~u ki~n (*) ~m=O nhan
+ Ne'u m "::F- 0 thl (3) nghiern duy nha't (l = 1 the' vao (4) ta dlf<;1c:
2m + 1 = 0 ~ m = - ~ thoa (*)
2
V~y: Cac gia tri cua m thea man yeu c!u bai toan ": m = 0 v m = -~
Ba] 8: Ghl sii' hai phirong trlnh:
a, x2 + bl X + cI = 0 va a2x2 + b2x + c2 = 0 c6 nghiern chung (It nha't mQt nghiem),
Chang minh rang: (aIc2 - a2cl)2 = (aIb2 - a2bI)(c2bI - cIb2)
Giai
Giii sii' Xo la nghiem chung cua hai phirong trinh da cho thl Xo thoa man h~:
{alx~ + b1xo + c1 ~ 0 (.1) ~ {a2aIx~ + a2bIxO + a2cI : 0 (3) a2xO +b2xo + c2 - 0 (2) aIa2xO + aIb2xO + aIc2 - 0 (4)
~ Xo (a2bl - aIb2) + a2cI :- aIc2 = 0 (*)(la'y (3)- (4»
Nhan (1) voi b2 va nhan (2) vdi bl ta dlf<;1C:
{al b2X~ + b, b2xO + cI bl = 0 a2bIxO + b2b1xO + c2b1 = 0
~ (a1b2 - a2bd x~ + c1b2 - c2bI = 0 (**)
Tir(*) ~ Xo = a1c2 -a2c1 the vao (**) ta diroc: (a2b1 -a1b2"::F- 0) a2bI - aIb2
(a1b2 -a2bl)[al~2 =a2~1]2 =c2bl -cIb2 a2 1 a1 2
~ (a1c2 - a2c1)2 = (c2bl - c1b2)(aIb2 - a2bl) (dpcm)
BM 9: Cho hai phirong trlnh:
ax2 + ( a - b) x + b = 0 (1)
va bx2 + (b - a) x + a = 0 (2)
Dinh cac cap s5 (a, b) sao eho hai phirong trlnh tren la nrong dirong nhau.
Gfai
1. Neu a = 0 thl: (1) ~ -bx + b = 0 (3)
50
(2) <=:> bX2 + bx = 0 (4)
a. Ne'u b = 0 thl (3) va (4) dung vdi '<Ix E R nen: .. {a = 0 thea d~. b=O
b. Ne'u b -:1= 0 thl (4) co nghiern [:: ~1 nhirng x = 0 khong thoa man d~ bai.
Vh c~p (0, b) vdi b -:1= 0 khong thoa man d~ bai.
2. Ne'u b = 0 thl (1) <=:> ax2 + ax = 0 ; (2) <=:> -ax + a :: 0
3.
Ly Iuan nhir tren ta cling diroc cap: {~ : o thea man d~.
, {a -:1= 0 2
Neu Ta c6: ~l = ~2 = (a - b) -4ab
b-:1=O
a. Ne'u (a - b)2 - 4ab < 0 <=:> a2 - 6ab + b2 < 0 <=:> [3b - 2J2b < a < 3b + 2J2b ne'u b > 0 3b + 2.J2b < a < 3b - 2.J2bne'u b < 0
<=:> 3 - 2.J2 < ~ < 3 + 2.J2 thl ca hai phirong trtnh d~u vo nghiem nen
chiing tu'dng dirong nhau.
{a -:1= 0 c: a rn
Suy ra c~p (a, b) vdi va 3 - 2,,2 < - < 3 + 2,,2
b -:1= 0 . b
[a ~ 3 +.2.J2
b N· '" b
. eu
a s 3 - 2.J2 b
thea man d~ toan,
(*)
GQi Xl; X2 lil hai nghiern cua phuong trlnh (1) xa; x4 IA hai nghiem cua phirong trtnh (2) E>~ (1) va (2) nrong dirong nhau thl:
{ a-b b-a
{Xl + X2 = Xa + X4. --a-:: --b- b
<=:> <=:> :: ±a
X1.X2 :: Xa.X4 a b
-=-
b a
Ke't hop vdi di€u kien (*) ~ b = -a
~ c~p (a, b) thea man Ia: (a;-a)vdi a E R
Ke't hop: Hai phirong trlnh (1) va (2) urong dirong khi va chi khi:
51
( a, b) = ( c, -a. ) v6'i a. E R
{a. :F- 0 va [3 :F- 0 (a,b)=(o.,[3)vO'i 3-2-12 < ~ <3+2-12
Bal 10: Xac dinh a, b, c sao cho phirong trlnh:
{bX2 + ex + a = 0 ax2 + bx + e = 0 (1) nrong dirong vdi M:
ex2 + ax+ b = 0
(2)
Giai
1. Ne'u phirong trtnh (1) co nghiern, d~ (1) va (2) urong dircng thl di~u kien dn
{aX2 + bx + e = 0
la: h~ bx2 + ex + a = 0 co nghiern, ex2 + ax + b = 0
Cong ve theo ve' ba phuong trlnh tren ta du'<;1c:
(a + b + e) x2 + (a + b + e) x + a + b +.c = 0
<=> (a + b + c)( x2 + X + 1) = 0
<=> a + b + e = 0 (vi x2 + x + 1 > 0 Vx E R)
=> b = -a - e. Til (1) ta du'<;1c: ax2 - x(a + e) + e = 0
<=> (x -1)( ax _ e) = 0 <=> [x == 1
. ' ax-e=O
{c = -a - b
Cling til: a + b + c = 0 => nen til (2) tao diroc:
a = -b-e
{bX2 - (a + b) x + a = 0 <=> {(X -1 )(bx - a) = 0 \. ex2 - (b + e) x + b = 0 (x -1)( ex - b) = 0
a. Neu a = b = c thl.(l) dung Vx E R nen (1) va (2) tu'dng dirong nhau.
b. Ne'u a2 + b2 + c2 > 0 (3 s6 a, b, c khong ddng thoi bang 0) Do a + b + c = 0 thl phai co it nha't 2 s6 khac O.
x=l
O· • • h . "" u {b:F- 0 hl , (2) {x = a "
ia sli' at so a t t ur <=> b
e:F-O b
X=e
+ Ne'u a = 0 thl (2) <=> x = 1 va (1) <=> x = 1 nen (1) va (2) tu'dng dirong nhau.
52
{a = 0; b :j; 0; c :j; O. ~
Suy ra: thoa man de bai.
a+b+c=O
[X = 1
+ Neu a :j; 0 thl (1) ¢:> x = ;.
a c b
Do do (1) va (2) nrong duong nhau ¢:> - = - = - ¢:> a = b = c rna
b a c
a + b + c = 0 => a = b = c = 0 vo 19.
[a = b = O'C:j; 0
2. Ne'u (1) vo nghiern ¢:> . ' 2
. a :j; 0 thi b - 4ac < 0
a. Tnrong hop a = b = 0; c :j; 0 thl tlt (2) ¢:> [cx 2= 0 ¢:> X = 0 cx = 0
Do d6 (1) va (2) khong nrong dirong nhau.
b. Truong hop a :j; 0 va b2 - 4ac < 0
+ Ne'u b = 0 thl (2) ¢:> t 2+ a = 0 ¢:> {[c: : ~ = 0
cx +ax=O 0
cx+a =
¢:> [{ :x =+ °a = 0 vo ly vi a :j; 0 ¢:> CX + a = O. cx +a e O
Phuong trinh nay luon c6 nghiem VI neu c = 0 thl b2 < 0 la di~u vo 19.
Do d6 (2) c6 nghiern nen tnrong hop nay (1) va (2) khong nrong duongnhau. + Neu b:j; 0 thl til' di~u kien b2 - 4ac < 0 => c :j; O. Khi d6 (2) vo nghiem.
c2 - 4ab < 0
¢:> a 2 - 2bc < 0
{C2 - 4ab ~ 0
a 2 - 4ab ~ 0 (3)
bx2 + ex + a = 0 va cx2 + ax + b khong co nghiem chung
{C2 -4ab ~ 0
H~ (3) a2 - 4bc ~ 0 (4)
b(b2 - ac)2 + c(b2 - ac)(c - ab) + a(c2 - ab)" :j; 0
T6m lai: (1) va (2) nrong dimng nhau khi va chi khi:
53
1. a=b=c=O
2. a + b + c = 0 trong d6 c6 hai s6 hang khac 0 va ml)t s6 hang bang 0
3. c2 - 4ab < 0 v a2 - 4ab < 0; a "i= 0; b "i= 0; C "i= 0
4. a"i= O;b"i= O;C"i= Ova thoa (4)
Bai 11: Gill sii' phtrong trlnh: ax2 + bx + C = 0 (a "i= 0) c6 dung ml)t nghiem
dirong la Xl' Chung minh rhng phirong trlnh: CX2 + bx + a = 0 cling c6 dung mdt nghiern dirong la x2. Chung minh rang: Xl + X2 ~ 2
Giai.
Phuong trtnh: ax2 + bx + C = 0 c6 dung ml)t nghiern dirong nen P = ac < 0 Tit d6 phuong trtnh: CX2 + bx + a = 0 cling c6 dung ml)t nghiern dirong.
Do Xl 1ft nghiern cua phirong trlnh:
ax2 + bx + c = 0 nen: axl2 + bXI + c = 0
Hon nii'a Xl > 0 nen: a + b J:_ + c ~ = 0
Xl Xl
~ J:_ la nghiern cua phirong trlnh: CX2 + bx + a = 0 (1)
Xl
Mat khac: Xl > 0 ~ J:_ > 0 rna phirong trlnh (1) chi co ml)t nghiem dirong Xl
nen nghiern dirong d6 la : X2 = J:_ 'XI
Vl v~y: Xl + X2 ~ Xl + _!_ ~ 2 (theo ba't dAng thtrc Cosi) Xl
~ Xl + X2 ~ 2 ~ dpcm.
Bai 12: Cho hai phirong trlnh:
X2 -12x+ m = 0
va X2 - 48x + q = 0
(1)
(2)
Gill sii' (1) c6 hai nghiern Xl; X2 ; (2) c6 2 nghiem xa; x4 .
a. Cac nghiern Xl' X2 ,Xa' X4 theo thu nr l~p thanh ca'p s6 nhan tang. TIm m, q.
b. Cac nghiern Xl' X2' Xa' X4 l~p thanh ca'p s6 cong. TIm m, q.
Giai
Phuong trtnh (1) c6 nghiemee tJ.' = 36 - m ~ 0 <=> m ~ 36 Phuong trtnh (2) c6 nghiern <=> tJ.' = 576 - q ~O <=> q ~ 576
54
{Xl + X2 = 12 {X3 + x4 = 48
Theo dinh 19 Viet: va
Xl·X2 = m X3.X4 = q
a. G9i cong bQi ciia c[p s6 nhan d6 la: d, theo tinh eMt ciia c[p s6 nhan ta c6:
X2 = dx.: X3 = d2Xl va X4 = d3xl "
{Xl + X2 = 12 {Xl + dx, =12 {Xd1 + d) = 12
Do d6: <:::> 2 3 <:::> (2 3)
X3 + X4 = 48 d Xl + d Xl = 48 Xl d + d = 48
d2 (1 + d)
=> =4=>d=±2
1+d
Do d~ cho la c[p s6 nhan tang nen ta chon d = 2.
{X2 = 2Xl {X3 = 4Xl
Voi d = 2 => va
Xl + x2 = 12 x4 = aXl
=> Xl = 4;X2 = 8;X3 = 16;X4 =·32
Suy ra: m = 32; q = 512
b. G9i P lA cong sai cua ca'p s6 cong thl:
{X2 = Xl +P
X4 - X3 = X3 - Xl = X2 - Xl = P => X3 = Xl + 2P X4 = Xl + 3P
{Xl + X2 = 12 {2Xl + P = 12 P
=> => => =9
x3 + x4 = 48 2Xl + 5P = 48
. { 63
3 21 39 57 m =·4
=> Xl = "2;x2 = "2;x3 = "2;x4 ="2 => 2223
. q=--
4
Bai 13: Cho cac phuong trlnh:
X2 + ax + 1 = 0 (1)
X2 + bx + 1 = 0 (2) X2 + ex + 1 = 0 (3)
Bie't r~ng mot nghiern ciia (1) vdi mQt nghiem nao d6 ciia phirong trlnh (2) lA nghiern cua phirong trlnh (3).
Chung minh rang: a2 + b2 + e2 + abe = 4
Giiii·
G9i Xl lA mot nghiem ciia phirong trinh (1)
x2 lA mQt nghiem cua phuong trinh (2)
55
Thl Xl' X2 Ia mot nghiern ciia phirong trinh (3)
Ap dung dinh Iy Viet eho phirong trlnh (1) ta diroc:
{Xl + X~ = -a 1
. => Xl +- =-a
Xl,Xl = 1 , Xl
Ttrong ttl d6i vdi phirong trinh (2) va (3) ta cling du'<;1c:
jx..2 + 2_ =-b
x2
Xl~ +_1_ =-c Xl~
U(y (4) nhan (5) v6 theo v6 ta diroc:
(4)
(5)
(6)
X X
=> _1 + __1_ = ab + c (Theo (6»
X2 Xl
Tit (4) va (5) ta du'<;1e:
( 2 ) (2 ) (2 1) (2 1) 2 2 1 x~ x~
=> a -2 b -2 = Xl +2 x2 +2 = XIX2 +~+2+2
Xl X2 Xl,X2 X2 Xl
=(XI~ +_1_)2 +(XI + ~)2 -4 =c2 +(ab+ c}2_4
XI'~ ~ Xl
~ (a2 _ 2)(b2 - 2) = c2 + (ab + C)2 - 4
~a2b2 _ 2(a2 + b2)+ 4 = c2 + a2b2 + 2abc+ c2-4 ~ a2 + b2 + c2 + abc = 4 (dpcm)
56
CHUYEN DE 7: CAC UNG Dl.JNG DIEU KleN co NGHItM CVA PHUONG TRINH n!c HAl
1. Chung minh ba't ding tlurc
Phuong phap: Til' mot dhg thirc T = f(x, y), ta bien d6i ve phirong trinh bac hai. Phuong trlnh nay eo nghiern <=> L\ ;?: 0 ~ dieu phai clnrng minh.
2. lim max; min eua ham s6 Phuong phdp
• Til' ham s6 y = f(x), dua v~ mot phirong trlnh b~c hai.
• Dung dieu kicn co nghiern cua phirong trlnh b~c hai de tim max, min.
3. Bili tiJP
Bili 1: TIm gia tri 1<'1n nha't va nho nhtt ciia ham s6:
) 6x - 1 b) . x2 - 2x + 3
a Y = Y = --:=----
x2 + 8 3x2 + 2x + 1
Giai
a. D=R
6x -1 2 .
Ta co: y = 2 <=> yx - 6x + 8y + 1 = 0 (1)
x + 8
[y = 0 [Y = 0
(l) co nghiern <=> t: 0 <=> {y:;t: 0 <=>
L\' ;?: 0 9 - y (8y + 1) ;?: 0
Til' do ta du'c;1c: maxy = 1 va miny = -%
b. D=R
x2 - 2x + 3 2
Ta co: y = 2 <=> (3y -l)x + 2(y + l)x + y - 3 = 0 (1)
3x + 2x + 1
3y-1=O
[3Y -1 = 0 (1) co nghiem <=> {3~ - 1 = 0 L\ ;?: 0
Til' do ta du'c;1C: maxy = 3 + 2J2 va miny = 3 - 212
Bal 2: TIm gia tri ldn nhtt va nho nha't cua bieu thirc: x2 - x + 1 .
y = 2 vdi x E R
x + x+ 1
57
Giii
2 (1 )2 3 A X2 - X + 1
Vi x +x+1= X+- +->0 vx e R nen: y=-2---
2 4 x +x s-L
<=> y (X2 + X + 1) = X2 - X + 1 <=> (y - 1) X2 + (y + 1) x + y - 1 = 0 (1) + Neu y = 1 thl (1) trCf thanh: x = 0
+ Neu y 'i:- 1 thl (1) co nghiern <=> A' ~ 0
<=> _3y2 + 10y - 3 ~ 0 <=> ! s y ~ 3 3
VA 3' 1 ay. max y e ;mmY=3
B"3 Ch harn sd x+ I D' h d" .~ '': . • ha ,..- hii [01]
81: 0 am so: y = 2 .!n a e mien gia tn eua am SO e U'a ,
X +a .
Giai
Di~u kien: X2 'i:- -a
GQi T la mien gia tri ciia ham s6, the tht t6n tai it nha't mQt s6 Yo E T sao eho
h ' h x + 1 , h'A
P irong trtn : Yo = 2 eo ng rem,
x +a
<=> yox2 + ayo = x + 1 <=> yox2 ~ x +ayo -1 = 0 (1)
1. Neu Yo = 0, tit(1)~ x = -1. Di~u kien X2 'i:- -a ~ a ~-1
2. Neu Yo 'i:- 0 thl (1) eo nghiern <=> A' = 1- 4yo (ayo -1) ~ 0 = 4ay~ - 4y 0 - 1 ~ 0
a. Ne'ua=Othl(2) ~Yo ~-! ehua{0,1]~a=0 thoa man de. 4
, , 1-vft+il 1+.J1+a., ( )
h. Neu a > 0 thl (2) <=> s y s (vdi A = 4 1 + a )
2a 2a /
D~ mien giatri nay chua [0,1] thl:
11-~' •
2a ~. 0 <=> {.J1 + .a ~ 1 hi~n nhien vi a > 0 .
1 + ~ ~ 1 .J1 -: a ~ 2a - 1
2a , .,
58
c. Neu a < 0
Neu a ::;; -1 tht t-.' = 2(1 + a) ::;; 0 ~ ba't phuong trinh (2) Iuon dung. Vdi VYo ~ a ::;; -1 thea man d~: .
N€u-l<a<O
D~ mien gia tri nay chua (0, 1) thl:
1 - ~1 + a tr:":
----::;;0~,,1+a ::;;1~a<0 (**)
2a
Tit (*) va (**) ta dtt<;1c gia tri ciia a dn tim la: {a s : a ~-1
Chii y: Bai nay ta dung phuong phap dao ham cling ra't tien loi. _
B' . 4 Ch h' ~ x2 - 2x + 4 - a al: 0 am so: y = --_,,----
x2 +a
Dinh a d~ mien gia tri cua ham s6 chua [0,1]
l__ . _
Giai
Di~u kien dn d~ bai toan dtt<;1c thea man Ia phirong trlnh:
y = x2 - 2x + 4 - a (1) phai co nghiern khi y = 0
x2 +a
{X2 - 2x + 4 - a = °
Khi Y = ° thl (1) trd thanh: 2 . (2)
x +a ~ °
Phuong trlnh (2) co nghiem ~ t-.' = 1 -: (4 - a) ~ ° ~ a ~ 3 Khi a ~ 3 ~ x2 + a > OVx
Ltic do: (1) ~ yx2 + ay = x2 - 2x + 4 - a
~ Ay _1)x2 + 2x + ay - 4 + a = ° (3) + N€u y = 1 thl: 2x - 4 + 2a = ° ~ x = 2 - a .
V~y y=1E[0,1]
+ Neu y ~ 1 thl (3) co nghiem ~ t-.' = 1- (y -1)(ay - 4 + a) ~ ° ~ ay2 - 4y - a + 3 ::;; °
59
Tam thirc f(y) = ay2 - 4y - a + 3 c6:
{f(O) = 3 - a ~ Ov6'i va ~ 3 f(1) = -1 < 0
Do d6 f(y) luon c6 hai nghiern Yl' Y2 thea: Yl ~ 0 s Yz M~t khac: f(y) s 0 <=> Yl ~ Y s Y2'
Do d6 [0. 1] chira trong mien gia tri cua ham s6 da cho <=> a >- 3 KC't luan: gia tri cua a dn tim ia a >- 3
each khdc
Di~u kien: x2 + a ::;:. 0
GQi T Ia mien gia tri cua ham s6, the' thl t6n tai it nhat mot s6 Yo E T de x2 - 2x + 4 - a
phuong trlnh: Yo = 2 co nghiern.
x +a
<=> (y 0 . - 1) x2 + 2x + ay 0 - 4, + a = 0 (1)
+ Ne'u Yo =1thl x=2-a=>yo E[0,1]
+ NC'u Yo ::;:.1 thl (1) c6 nghiern <=> tJ.' = 1- (Yo -1)(ayo - 4 + a) ~ 0 <=> ayo 2 - 4yo - a + 3 ~ 0
Ne'u a = 0 thl : -4yo + 3 s 0 => Yo ~ ~ khong chua he't [0,1] 4
a.·
=> Ne'u a = 0 khong thea man d~.
• Ne'u a < 0, tam thtrc f(yo) = aYo2 - 4yo - a + 3 co
, ( 3)2 7 '
tJ. = 4 - a(3 - a) = a -"2 + 4' > 0 'v'a => tJ. > 0 va
Do d6: f(y) ~ 0 <=>
2+~a2-3a+4 0\-1 0
Yo s < va <
a
2....;~a2 -3a+4 Yo~-----a
• De mi6n gia trj cua ham s6 chira [0. 1] thl:
2-~a2 -3a+4 ------~O
a
<=> ~ a 2 - 3a + 4 ~ 2 <=> a 2 - 3a + 4 ~ 4 <=> 0 < a ~ 3
KC't hop vdi di~u kien a ~ 0 suy ra trong tnmng hop na ykhOng tlm duc;fc a thea man d~ bai.
b. Ne'u a > 0, tam thirc f(yo) s 0
60
2-Ja2 -3a+4 2+Ja2 -3a+4
¢::> 5, Yo 5, ------
a a
D~ mien gia trj nay chua [0, 1] thl:
2 - Ja2 - 3a + 4
a 5, 0 {Ja2 - 3a + 4 ~ 2
¢::> ~oa>m
2 + Ja2 - 3a + 4 ~ 1 Ja2 - 3a + 4 ~ a - 2
a
a2 - 3a + 4 ~ 4
¢::> [a - 2 5, 0 ¢::>
{a- 2 ~ 0
a2 - 3a + 4 ~ (a - 2)2 = a2 - 4a + 4
T6m lai: gia tri cua a c~n tim la a ~ 3 .
Bai 5: Tim gia tri ldn nha't va nhO nha't cua ham sf): 3rx+3+4~+1
Y=
4../x + 3 + 3../1- x + 1
Giai
Di~u kien: -3 5, x 5, 1
Do do: ( ../X + 3 f + ( ../1 - x t = 4 nen ta co th~ d~t:
I../X+3=2~
1+t '·0< <1
2 vdi - t -
~=2~
1 + t2
Khi d6: ham s6 dii cho trCf thanh: _7t2+12t+9
Y = --;;-----
-5t2 + 16t + 7
¢::> Y (-5t2 + 16t + 7) = -7t2 + 12t + 9
¢::> (5y -7)t2 + (12 -16y)t + 9 -7y '= 0 (1)
+ Ne'u 5y -7 = 0 ¢::> y = 2_ thl (1) trCf thanh: t = _ _l ~ [0,1]
5 13
N'" 7
+ eu y '# '5:
D~t: f(t) = (5y -7)t2 + (12 -16y)t-+,9 -7y Dinh y d~ f(t) = 0 co nghiem t E [0,1]
Xet cac tnrong hop:
61
[0 s t < 1 s t
a. f(t) = 0 co 2 nghiern t1; t2 thoa: 1 2
tl < 0 s t2 < 1
7 9
~ f(O).f(1) s 0 ~ - ~ y ~ -
9 7
b. f(t) = 0 co 2 nghiern t1; t2 thea man: 0 <: tl s t2 < 1
/).' ~ 0 7 9
-<y<-
(5y - 7) .f(O) > 0 9 7
14 7
~ (?y - 7) .f'(I) > 0 ~ -<y<-
8 5
0< 16y-12 <1 0< 16y-12 < 1 va /).' ~ 0
2(5y -7) 2(5y-7) ~
8y- 6 < 1
5y-7
7 9
-<y<-
9 7
(8y-6)(5y-7) > 0
~
7 9 -<y<- 9 7
[Y< ~
y>-
5
8y- 6 < 1 5y-7
/).' ~ 0
VQ nghiem
Tom lai:
7 9 {maxy = ~
_ ~ y ~ _ nen: 7
9 7 . 7
mmY=g
B' . 6 11 . ..:-. ld h~ " h" h~ • h" 2 + sin 2x
31 : im gra tn on n at va non at cua arn : y = 2
2 + cos x
Ta co: y = 2 + si~ = _---,C=O=S2---,X,=,"-_ = _2t-=g:...2_X-:+:-2_;tgx~+_2
2 + cos'' X 2 + cos2 X 2tg2x + 3
Giiii
2 + 2 sin x. cbs x
cos" x
2e + 2t + 2 2
D~t:t=tgx,tadu'~c: y= 2t2+3 ~2(y-1)t -2t+3y-2=O (1)
+ Ne'u y::: 1 thl (1) trd thanh: -2t + 1 = -0.::::> t = .!.
. . 2
+ N6u y 'j:. 1 thl (1) co nghiern
62
. 5-.fi 5+.fi
~~ ~o~ ~y~---
6 6
j. 5-.fi
mIny=---
Tit d6 suy ra: 6
5+ .fi
maxy = 6
Bal 7: Cho cac sf) x, y thoa man: x2 + y2 + xy = 1
TIm gia tri ldn nha't va nho nha't ella: A = 2X2 - xy +3y2
Giai
2X2 - xy + 3y2 Ta e6: A = 2X2 - xy + 3y2 = -::-------=:----"-x2 + y2 + xy
+ Neu y = 0 thl: A = 2
( )2
X X
2. - --+3
+ Neu y * 0 => A * 2 khi d6 A = Y 2 Y
(x) x '
y +y+l
DAt t = x thl A = 2t2 - t + 3
. Y t2 + t + 1
(1)
C~n xac dinh A d~ (1) e6 nghiern:
~ ~ ~ 0 ~ 11-$2 ~A~ 11+$2
3 3
jmaXA = 11 + $2
Tit d6 suy ra: 3 r;::-;:;
. A 11-,,52
mIn =---
3
Bai 8: Cho 2X2 + y2 + xy ~ 1.
TIm gia tri nho nha't ella bi~u thtrc: M = x2 + y2
Giai
M 2 2
D~ t: 2x2 + y2 + xy = a vdi a ~ 1 . Khi do - = 2X +-;
a 2x + y + xy
+ Neu y = 0 tht: M = .! a 2
63
(~r +1
+ Ne'u y -:I; 0 thl: M = -__:"':'_-':-2---
a 2.(~) +~+1
D<\t: t = x va A = M thl:
. y a
e +1
A = 2 <=:> (2A - 1) t + At + A - 1 = 0
2t + t + 1
Phuong trinh nay co nghiern: <=:>~~O
<=:> 6 - 2J2 s A s 6 + 2J2
7 7
=:> M ~ 6 - 2J2 =:> M ~6 - 2J2 .a ~ 6 - 2J2
a 7 7 7
=:> Min(M) = 6 - 2J2
7
VA • , • h? h~ ? M I' 6 - 2J2
ay gra tn n () n at cua a: ---
.. 7
Bai 9: Chung minh ding vdi va, b, c, d, e E R ta luon co:
a2 + b2 +c2 +d2 +e2 z ab e ac e ad t ae (1)
J
Giai
Bat d&ng thirc: (I) <=:> a2 + b2 + c2 + d2 + e2 - a(b + c + d + e) ~ 0 X6t tam thirc h~c hai:
f'(x) = a2 -(b+c+d+e)a+ b2 +c2 +d2 +e2
Ta co: ~ = (b + c + d + e f - 4(b2 + c2 + d2 + e2) Theo hat dftng thirc Bunhiakopxki:
(b + c + d + e)2 s 4(b2 + c2 + d2 + e2)
=:> (b + c + d + e)2 - 4(b2 + c2 + d2 + e2) s 0 =:> ~ :s; 0 =:> f'(x) ~ 0 \fx E R =:> (dpcm)
Bal 10: Cho tam giac ABC. Chung minh r~ng:
1 . A . B . C 1 .. sm-.sm-.Sln-:S;-
2 228
2. cos A + cos B + cos C :s; ~ 2
3. sin2A+sin2B+sin2C:s;~ 4
64
1.
Giai
.A .B'.C 1 S·A.B.C 1
Sln- .Sln-.SIn-:::;; - ~ Sln-.SIn-.SIn-:::;;
222 S 222
Xet hi~ll thtrc:
T S· A . B . C 4' C ( A- B A+ B)
= Sln- .Sln- .Sln- = Sln- cos-- - cos--
2 2 2 2 2 2
T 4· C ( A- B . C)
~ = Sln.,.- cos-- -Sln-
2 2 2
4 . 2 C 4 A- B . C T 0
¢:> SIn - - COS-- .Sln- + =
2 2 2
Xcm phirong trlnh na y la phirong trlnh h~c hai thea ffn u sin ~
Phuong trlnh nay co nghiern ncn: !!.. > 0
¢:> 4 cos2 A - B _ 4 T ~ 0 ~ T :::;; cos'' A - B s 1
2 2
. A . B . C 1 (d ).
~ sm2":sm2.sm2:::;;"8 pcm
2.
f)a t T = cos A + cos B + cos C = 2 cos A + B . cos A - B + 1 _ 2 sin 2 C
. 2 2 2
2 . 2 C 2 A-B. c T 1 0
~ sm -- COS--.SIn-+ - =
2 2 2
Xem phirong trlnh nay 13. phirong trlnh b~c hai thea ffn la sin ~
Phuong trlnh nay phai c6 nghiem nen: !!.. ~ 0
¢:> cos" A - B_2 (T _ 1) ~ 0 ~ T :::;; 1 + .!. cos" A - B:::;; 1 + .!. = ~
2 2 2 2 2
3
~ cos A + cos B + cos C ,:::;; "2 ( dpcm)
• 2 • 2 • 2 1 - cos 2A 1 - cos2B 2
f)i\ t: T = sin A + sm B + sm C = + + 1 - cos C
, 2 2
1
== 2 - - (cos 2A + cos 2B) - cos2 C 2
= 2 - cos( A + B) cos( A - B) - cos" C
= 2 + cosC.cos(A+ B) - cos2 C
¢:> cos" C - cos( A - B) cosC + T - 2 =, 0
Xem day la phirong trlnh b~c hai thea ffn s61a cosC.
3.
65
Phuong trlnh nay phai c6 nghiem nen:
L\ ~ 0 <=> cos2 (A - B) - 4 (T - 2) ~ 0 1
~T~2+ 4 cos2(A-B)
~ T s 2 + 1. = ~ ~ sin 2 A + sin 2 B + sin" C s 9
4 4 4
CHlIdNG 2. PHUdNG TRiNH B~C BA, B~C BON vA DA THaC
A. LY THUY~T CAN NHa:
I. Nghi~m ciia da thdc:
1. siS c la nghiem bQi Ie ella mQt da thrrc Pn (x) ne'u Pn (x) chi a he't cho
(x - c t vdi k Ie. Ne'u k = 1 thl c goi la nghiem don,
2. se' c la nghiern bQi chin efta da thtrc Pn (x) ne'u Pn (x) chia he't cho (x - e t
va k la siS chin. Ne'u k = 2 tht c la nghiem kep.
3. MQt da thtrc b~c n chi c6 th~ c6 tiSi da n nghiern.
4. MQt da thirc b~c Ie luon c6 it nha't mQt nghiern.
5. MQt da thtrc b~c n, rna c6 (n + 1) nghiern thi da thirc d6 d6ng nha't khOng.
II. Dinh It Viet ap dung cho phttdng trlnh b~c ba:
Cho phirong trlnh b~c ba: ax' + bx? + ex + d = 0 (a * 0). Giii sil' phirong trlnh
na y c6 ba nghiem XI' x2, X3 • Khi d6 ta c6 he; tlnrc:
b
XI + X2 + X3 = -a
m. each nhAm nghiern nguyen ho~c nghi~m hiYu ti cua phttdng trlnh b4c ba
• Ne'u plnrong trlnh b~c ba c6 th~ c6 nghiern hiru tl hoac nghiern nguyen, d~ tlm nghiem nay ta lam nhu sau:
+ TIm ~p hop cac u'dc siS ella a goi la p.
66
+ Tim t~p hop cac u'oe sCi cua d goi Hi q.
+ Tim cac phan sCi q ; r6i the' tung phan sCi vao phirong trlnh; pharr sCi nao p
thea man tht do Ia nghiern ciia phirong trlnh,
B. CAC CHUYEN DE TOAN vA PHU<1NG PHAP GIAI
,. ~
CHUYENDE 8:
AI,.. A ~ . ",
NHAM MQT NGHI~M - DUA VE PHUONG TRINH TICH
Ba! 1: GhH cac phirong trlnh sau:
1. 2x3-5x2+3x+l0=0
2. x3 -3x+2 = 0
(1) (2)
Giai
1. Ta tha'y phirong trlnh (1) co mQt nghiern x =-1 Dung so d6 Horner dira ve tich:
(1) <=> (x + 1)(2x2 -7x + 10) = 0
[X = -1 <=> x = -1
<=> 2X2 -7x + 10 = O(vo nghiem )
2. Phuong trlnh (2) <=>(x-l){x2+x-2)=0
[X -1 = 0 [x = 1
<=> x2+x-2=0 <=> x=-2
Bai2: Cho phuong trlnh: x3-(2m+l)x2+mx+m=0 (1)
Tim m d~ phirong trlnh co ba nghiern phan biet,
Giai
(1)<=>(x-l)(x2-2mx-m)=0 <=>[x=1 2
f(x)=x -2mx-m=0 (2)
f)~ (1) e6 hai nghiern phan biet thl (2) co hai nghiern phan bi~t:;:. 1
{L1' > 0 {m2 + m > 0 [m < -1
<=> <=> <=> 1
f(1):;:.O 1-3m:t0 O<m:;:'3"
Bai 3: Cho phirong trlnh: x3 -2(m + l)x2 +(7m - 2)x +4-6m = 0 (1) Tim m d~ phunng trlnh e6 ba nghiern dirong.
67
Giiii (1) ¢::> (x -2)( X2 -2mx +3m-2) = 0
[X =2
¢::> f'(x) = x2 - 2mx +3m- 2 = 0 (2)
J)~ (1) c6 ba nghiern phan biet duong thi (3) c6 hai nghicrn phan bi9t dirong va khac 2
f(2) ~ 0
l~<m<1
¢::> 3
m>2
Yay cac gia tri cua m dn tim la: ~ <m < 1 v m > 2
.. 3
Ba] 4: Cho phirong trlnh: mx ' -( m _4)X2 +(4 + m) x -m = 0 (1) TIm m d~ phirong trinh co dung hai nghiern.
Giai
(I) ¢::> (x-I)(mx2 -4x+m)=0
[X = 1
¢::> f(x)=mx2-4x+m=0(2)
• Vdi vrn E R thi (1) lunn c6 mot nghiern x = 1
• J)~ (1) co dung hai nghiern thl (2) c6 dung mot nghiern ~ 1.
Xet cac tnrong hop sau:
THl: Neu m = 0 thl (2) ¢::> x = 0 ~ 1 => gia tri m = 0 thea d~ bai.
TH2: Neu m ~ 0 (2) c6 nghiern kep ~ 1
{f(1)~O {2m-4~0
¢::> _ ¢::> 2 ¢::> m = -2
~ =0 4-m =0
TH3: Neu m ~ 0 (2) c6 mot nghiern XI = 1 con nghiern x2 ~ 1
{~. > 0 {4 - m2 > 0
¢::> ¢::> ¢::> h~ va nghiern
f(I)=O 2m-4=0
Vh cac gia tri ciia m dn tim la: m = 0 v m = -2
68
ri 5: Cho Phlfdng. trlnh: x ~ +(1- m) X2 - 3mx + 2m2 = 0 (1) TIm m 00 phlf(fng trinh co oung m(>t nghiem.
---_._--_._------_ .•.. _-, .. -_
Giai
[x=m
(1) <=> (x -- m) ( x 2 + X - 2m) = 0 <=> 2
f'(x) = X + X - 2m = 0 (2)
• V di \1'111 E R thl (I) iuon co mot nghiern x = m
• D( (I) co dung rno I nghcrn thi La ,\.6L eric tnrong hop sau: 1 TH I· (2) \\., 1~~hiGO' ~ .. L\ < 0 <=> 1 + 8m < 0 <=> m <--
, . 8
Ti-l2: C) u) nghiern kcp '; = m
f f1 = () {I + 8m = 0
<=> '1 ' <=» <=> he vo nghiern
[f'(rnj v O m--m=O
,\ 1
V,\V de gi;i tri cua m can tim 1a m <-_
.J. 8
lB~i6: Cho phuong trlnh: x3-6x2+(5m+l)x-3m=0 (1)
LTIm m otJ phu(1ng trlnh co 3 nghiem phan biet 1~p thanh d(p Sel e(l_n.=..g_. --'
Giai
Ctich I: Dung dinh 1y Viet:
Giii sli (I) co 3 nghicrn phan bi(:t XI < Xl < x:1l~p thanh cAp s6 cong:
{XI+XJ+X1=6 (2)
Thco dinh Iy Viet ta co: _.
XI+x1=2x2 (3)
Tu (2), a (3) ~ x2 = 2 the vao (1) diroc: m = 2
• Khil1l:.-;2lhl(l) <=>x3-6x2+11x-6=0
<=> r:~ : ~
X3 =.3
Ta Iha'y ha nghicrn nay l~p thanh cAp Sel cong. V~y III = 2 la gin tri dn tim.
C(Ich 2: Dung tinh chAt di&m u6n cua ham bac ba:
"Cho ham :-.6" b~c ba: f'(x) = ax) + bx ' + ex + d = O( a :t: 0).
Df> th] ham b~c ba dt Ox tai ba di6m phan biet co hoanh dO 1~p thanh cAp s6 c(lIlg
69
¢:::>{f' (x) =" 0 c6 hai nghiem phan biet di€m u6n thuoc Ox
D~t f'(x) = x ' -6x2 +(Sm+l)x-3m f'{x) = 3x2 -12x +Sm + 1
((x) = 6x-12;((x) =0 ¢:> x = 2 Yell ch bai toan du'<1c thoa man
¢:::> {f' (x) = 0 c6 hai nghiern phan bi~t f(2) = 0
{i\' > 0 {36 - 3 ( 5m + 1) > 0
¢:> ¢:> ¢:> m = 2
7m-14 m=2·
V~y m = 21a gia tri c~n tim.
each 3: Thi€t l~p bQ nghiern l~p thanh ca'p so' cong.
Gid sit tim tai so' a > 0 sao cho ba so' a < 2a < 3a l~p thanh ca'p so' cong cua phirong trlnh ban d~ u.
Khi d6: x3 -6x2 +(Sm+l)x -3m =(x-a)(x-2a){x-3a)
=> x ' -6x2 +(Sm + l)x -3m = x3 -6ax2 + 11a2x -6a3 (1)
D6ng nha't h~ so' hai v€ clla (1) ta du'<1c:
{~:a~ : ~~ + 1 ¢:> "{a = 1 => m = 2 "
m=2 -6a3 =-3m
Thit lai: Khi m = 2 thl phirong trinh da cho:
¢:> x ' -6x2 +llx-6 = 0 ¢:> XI =1 V x2 =2VX3 =3 Ro rang ba nghiern nay l~p thanh ca'p so' cong.
V~y m = 213. gia tri c~n tim.
CHUYEN DE 9: DUNG PHUONG PHAP KHAO SAT HAM s6
1. Dinh Iy:
Phuong trlnh f'(x) = gem) c6 nghiem XED khi va chi khi g(m) nAm trong mien gia tri cua f(x).
2. Cac bd8e giai:
• Buac l: Dira phirong trlnh v~ dang; f(x) = g(m) (1)
70
• Blltlc 2: f(x).
• BLMc 3: Dira vao dinh Iy tren d~ tlm m.
L4p bang bien thien ciia ham f(x) d~ tim mi~n gia tri ciia ham
3. Chu y:
a. Phuong trlnh (1) c6 n nghiern thl du'Cfng th£ng y = g(m) ci(t d6 thi ham f(x) tai n di~m.
b. Phuong phap nay ap dung khi khong nhim du'<;1c nghiern,
4. BM ttp
Bai 1: Giai phuong trtnh: x3 - 3x2 + 1 = 0 vdi 1 < x < 2
Giai Xet ham f(x) = x3 - 3x2 + 1
f'{x) = 3x2 -6x;f'(x) = 0 <=> x = Ovx = 2
Bang bien thien:
2
Theo bang bitn thien: => f'(x) < -I V'x e (1;2) => phirong trlnh dii cho vo nghiem,
I BM 2: Giai phirong trtnh: x3 -3x + 1 = 0 (1)
Giai
• Ta chang minh (1) luon c6 3 nghiern x e [ -2,2] Cach 1: dung dinh Iy lien tuc:
D~t f'(x) = x3 - 3x + 1 la ham so' lien tuc tren [-2, 2]
Ta c6: f(-2) = -1; f(-I) = 3; f(1) = -1; f(2) = 3 =>f(-I).f(I)<O=> (Jj luon co mot nghiern xe[-2,2]
=>f(-2).f(-I)<O=> (1) luon co mot nghiern xe[-I,I] => f(I).f(2) < 0 => (1) luon c6 mQt nghiem x e [1,2]
=> phirong trlnh dii cho c6 ba nghiern phan biet x e [-2,2]
Cdch. 2: Dung khao sat ham so':
Xet ham so': f(x) = x3 -3x +1 vdi --2 ~ x ~ 2
f'{x) = 3x2-3;f'(x) = 0 <=> x = ±t Bang bitn thien:
71
x
o + 0
-2 -I 1 2
f(x)
f(x)
_____. 3 3
-I ~ -I ~
Ti't bang biC'n thien ~d6 thi ham f(x) luon dt true Ox tai ba di6m phan biet c6 hoanh d9 x e [-2,2]
• Gid i phirong trinh (I)
D~t x = 2cost; (O:s; t :S; JZ")
(I) tra thanh:
[ 2JZ" k2JZ"
o :S; - + -- :S; JZ"
Vi 0 :S; t :S; JZ" nen 9 3
2JZ" . k2JZ"
O:S; --+--:S; JZ"
9 3
8 cos' t - 6cost + 1 = 0
<=> 2 ( 4 cos" t - Jost ) = -I <=> cos 3t = - ~
[2JZ" [2JZ" k2JZ"
3t =-+k2JZ" t =-+--
<=> 3 <=> 9 3 keZ
3t = _ 2JZ" + k2JZ" 't:::: _ 2JZ" + k2JZ" ( )
3 3 3
[17
-_ -:s;k:s;-~k=Ovk=I
<=> 3 6
1 11
-:s;k:s;-~k=l
3 6
2JZ" 2JZ"
t = - ~ x = 2 cos-
9 9
SJZ" SJZ"
~Banghi~mla t=-~x=2cos-
9 9
4JZ" 4JZ"
t = - ~ x = 2cos-
9 9
o 0, 2 2JZ" 2 SJZ" 2 4JZ"
Vaybanghlcmcua(l)IA: x= cos-; x= cos-; x= cos-
o 0 9 9 9
Bai 3: Cho phirong trlnh: x' - 4mx2 + S = 0 (1) Tim m d6 phinrng trlnh c6 ba nghiern phan biet,
Giiii (l)<=>x3+8=4mx2 <=>x+-;'=4m (x s O) x
S .
Xet ham f'(x) = x +-2 vdi x;t 0 x
72
['(x)=I-.!2.= x3-16
X3 x3
Bang bie'n thien:
f'{x) = 0 ¢:> x =:J16
x -00 0 ~ +00
f(x) + 0 +
/,00 +00 +00
f(x) ~~/
-00 ~256 Theo bang bie'n thien ~ yeu du bai toan dtroc thou man.
24 6
¢:> 4m > -- ¢:> m >--
ifi56 ifi56
x3
Bal 4: Cho phirong trlnh: - - x2 - 3x + 3m + 4 = 0 (1)
3
TIm m de phirong trlnh c6 dung hai nghiern pharr biet Ion hon 1.
Giiii
x3
(I) ¢:> -- x2 -3x + 4 = -3m
3 .
x3
Xet ham s6: f(x) = ---'- - x2 - 3x + 4 vdi x > 1
3
[' (x) = x2 - 2x - 3; f" (x) = 0 ¢:> x = -1 v x = 3 Bang bien thicn:
-1 +00
+00
~-5/
Theo bang bien thicn ~ yeu du bal toan ou(.1c thoa man.
1 1 5
¢:> -5 < -3m < - ¢:> -- < m <- 393
Rai 5: Cho phirong trinh: x ' + 3x2 + (3 - 2m) x + m + 1 = 0 (1) TIm m d6 phirong trlnh co dung mot nghiern x > I
Giai
( ) 3 1 (x + 1)'
1 ¢:>x' +3x-+3x+l=m(2x-l) ¢:>. =m (vi x > 1 ncn 2x+l:;tO)
. 2x-l
(x+lf
Xct ham s6: f(x) = voi x > I
2x -I
73
( )2 ( )
, x +I 4x-5, 5
f (x) = 2; f (X) = 0 <=> m =-
(2x-l) 4
Bang bien thien:
1 5
x +00
4
rex) 0 +
8 +00
f(x) ~243 /
32 Theo bang bien thien ~ yeu du bai toan du'<;1c thoa man
[m= 243
<=> 32
m z S
eHUYEN DE 10: DUNG TINH CHAT eve TRI eVA HAM B!e BA
Cho phtrong trlnh b~c ba: ax" + bx ' + ex + d = o( a:;t 0) (1)
1. Phuong trtnh (1) c6 ba nghiem phan biet
¢:> {f' (x) = 0 c6 hai nghiern phan bi~t Xl; x2 f'(x, ).f(x2) < 0
2, Phuong trlnh( 1) c6 ba nghiern phan biet ldn hon s6 a cho tnrdc:
{nx) = 0 c6 hai nghiern phan biet thea man a. < x, < x2 ¢:> f'(x, ).f(x2) < 0
af'(u) < 0
3, Phuong trlnh(l) c6 ba nghiern phan biet ldn hon s6 a cho tnrdc:
{f' (x) = 0 c6 hai nghiem phan bi~t thea man Xl < x2 < a. ¢:> f'(x, ).f(x2) < 0
af'(n) > 0
4. Phuong trinh (1) c6 nghiern duy nha't:
[nx) = 0 vo nghiern .
¢:> {f' (x) = 0 c6 hai nghiem phan biet Xl; x2
f'(x, ).f(x2) > 0 .
5, Phuong trtnh (1) c6 ba nghiern l~p thanh ca'p s6 cQng.
74
, <=> {f '(x) = 0 c6 hai nghiem phan bi~t Xl; X2 di~m uO'n thuoc Ox
6. Bai tip
Bal 1: Cho phuong trlnh: x3 - 3mx + m2 = 0
TIm m d€ phirong trlnh c6 ba nghiern phan biet.
Giai
D~t f(x) = x3 - 3mx + m2
. 2 . -[XI =rm
f(x)=3x -3m;f(x)=O<=> x2 =_rm(m>O)
f'(x.) = f(rm) = m2 -2mrm
f(x2) = f(...:.rm) = m2 +2mrm
Yeu ch bai toan du'<Jc thoa man:
<=> {m > 0 <=> 0 < m < 4
f'(x, ).f(X2) < 0
V~y cac gia tri cua m c~n tim la: 0 < m < 4
Ba] 2: Cho phuong trlnh: x3 - mx + 1 = 0 (1)
1. Dinh m d€ phirong trlnh c6 dung ba nghiern phan biet.
2. Dinh m d€ phu'dng tdnh c6 dung ba nghi~m phan bi~t l~p thanh ca'p so' cQng.
Giai
1. Dinh m d€ phuong trlnh c6 dung ba nghiern phan biet,
Phuong trlnh (1) <=> x3 + 1 = mx <=> x2 + _!_ = m (x;t: 0 X
Xet ham so' f(x) = x2 + _!_ vdi x;t: 0 . x
. 2x3 -1 . ( ) 1
f (x) = 2 ; f x = 0 <=> x = .t:
x {l2
**Bang bien thien
**
0 1
x -00 ~ +00
rex) 0 + f(x)
75
DVa vao hang hiC'n thien suy ra: phirong t r Inh (2) co ba nghiern phan bict: 3
~m>-
if4
,
2. Dinh m 06 phirong trinh co dung ba nghicrn phan biet l~p thanh dip s6 Cl)ng.
Caelt 1:
G9i x1,x2,x,(theo tlur t~t 0(» la ba nghicrn phfi n bi¢t ella (I), l~p tha nh dip
s{') cong. Theo tinh cha't dlp sf) cong va dinh ly Viet ta co:
x1+x,=2X2 {
x1+x,=2x2
XI+X2+X,=0
~ X, =0
X1X2 + X2X, + X,X1 = -111 - _
X1X2 + x2XJ + X.lx1 -- -m x I .X 2 .X, = -1
Hi; vo nghicrn.
V~y khong t6n ta i giri tr] m d6 tho a 06.
Caeh2: f(x)=x3~mx+1
(x) = 3X2 -m; (x) = 6x;(x) =0 ~ x = 0
£><1 phirong trlnh ftx) = 0 co ba nghiern phan bict l~p thanh d(p s(; cong thi:
{r' (x) = 0 c6 hai nghiern phan biet f(O) = 0
~ {3X2 - m = 0 co hai nghiem phan biet 1=0
Hl; nay vo nghiern. V~y khong l{~n tai gia tr] m thoa 06.
Bili 3: Cho phimng trlnh: x ' - X 2 + 18ax - 2a = 0
£>inh a oe phu'(1ng ldnh co ha nghi¢m phan bi¢l Jl({~l1g: . _
----
Giai
Cach 1:
£>~t: f(x)=x3-x2+18ax-2a . (x)=3x2-2x+18a Phuong trlnh oil cho c6 ba nghicrn pha n hict duong:
{f<X) = 0 co hai nghiern phan biet : 0 < XI < x~ (I)
~ f'(x, )f(x2) < 0 (2)
f(O) < 0 (3)'
. .
£>i6u kien (I) ~ 3x 2 - 2x + 18a = 0 co bai nghicrn phf 11 bictduong:
76
• Di0u kien (3) <=> -2a < 0 <=> a > 0 U(y f(x) chia cho f'{x) ta diroc:
[( x) = ( ; - i) ( (x) + 2 ( 6a - i ) x
VI Xl; X2 la hai nghiern cua phirong trinh f{x) = 0 nen (XI) = (x2) = °
[(XI) = 2( 6a-i )X'I
=-:> hai gia tri eire tri:
{o <a <~
Tit (l) (2) (3) ta diroc: 4 vo nghiern
a<O
V:iy khong ton tai gia tri a thea man d~.
( tich 2:
( 101 o.;l't pilltdng trtnh orr cho co ba nghiern duong phan biet la: xl> X2' X3
{XI + x2 + x) = 1
! hco dinh Iy Viet ta co: ~IX2 + X~3 + X)XI = 18a XI.xl.X) - 2a
Thco ha't ohg thirc Cosi:
1 = XI + x2 + X~ ~ 3/xIX'X~ => 1~27.2a => a ~ _1 (1)
. 'J -. 54
18a = XIX2 + X2X3 + X3X1 ~ ~( XIX2X3)2
<=> (lSa f ~ 27.4a 2 => a ~_1
54
Tit (1) va (2) => a =_1 54
77
Mat khac VI ba nghiern Xl' X2' Xa phan biet nen a * 514 Do d6 khong t&n tai gia tri a thea d~.
CHUYEN DE 11: NGHItM CVA PHU<1NG TRINH n!c BA THOA MAN MOT DIEU KIEN CHO TRUDe
1. Phuong phap giai
• Birdc 1: TIm di~u kien cua tha~ s6 d6 phuong trinh co ba nghiern phan biet.
b
Xl +X2 +X3 =-a
• Birdc 2:
Dung dinh 19 Viet:
d
X1·X2·X3 ::--
a
~=============================---'-=-================~
2. Bai t~p:
Bai 1: Cho phirong trlnh: X3 + X2 - (m + 2) X + m + 1 = 0
1. Dinh m d6 phirong trlnh e6 ba nghiern phan biet Xl' X2' Xa thoa man:
Xl < x2 < 2 < X3
2. Dinh m d6 phuong trlnh e6 ba nghiem phan biet l~p thanh ea'p s6 cong.
Giai
1. f)i~u kien e~n:
f)~tf(x)= X3 +X2 -(m+2)x+m+l
Gid sii' f(x) = 0 e6 ba nghiern: Xl' X2' Xa thea: Xl < x2 < 2 < xa Khi d6: f'(x) = (x - xl)(x - x2)(x - x3)
va f(2)= (2-xl)(2-x2)(2-x3)<0 (do Xl <X2 <2<xa)
¢::> 23 + 22 - (m + 2) 2 + m + 1 < 0 ¢::> 9 - m < 0 ¢::> m > 9 f)i~u kien du:
Gia sir m > 9, khi d6: f(2) = 9 - m < 0 Lai e6: f(O) = m + 1 > 0 vm > 0
~ f(0).f(2) < 0 ~ f'(x) = 0 e6 mQt nghiem E (0,2)
VI lim f'(x) = +00 nen t6n tai s6 a> 2 rna f(a) > 0
x-++~
:::) f(a).f(2) < 0 ~ phuong trlnh f(x) = 0 e6 nghiem X3 thea man: 2 < X3 < a
78
vi lim f'(x) = -00 nen t6n tai s6 b < 0 sao cho f(b) < 0
x-+_·r.
~ f(b).f(O) < 0 ~ phirong trinh f(x) = 0 c6 nghiern Xl thea man: b < Xl < 0 NhU' vh phuong trlnh da cho c6 ba nghiern Xl' X2, X3 thuoc ba khoang khac nhau v~ thoa man: Xl < x2 < 2 < X3
V~y gia tri cua m c~n tim la: m> 9
2. GQi ba nghiern cua phirong trlnh da cho lA Xl' X2' X3 ·theo thu tt! d6 l~p thanh
,.., ,.., A
cap so cong.
D~t: f'(x) = X3 +X2 -(m+2)x+m+1
f" (X) = 3X2 + 2x - m - 2
" " 1
f (x) = 6x + 2; f (x) = 0 <:::> X = --
3
Yeu du bai toan dU'c;1C thea man
{£' (x) = 0 c6 hai nghiem ph an biet 47
<:::> (1) <:::>m=--
f -- = 0 36
3
VA 47 1" . .e • ~ ,
ay m = - - 4 gl4 tn can 11m.
. 36 .
Bai 2: Cho phuong trinh: X3 + ax + b = 0 . Giii sit phirong trinh c6 ba nghiern la ~ Xl' X2' X3 . Chung minh rhng: x~ + x~ + x~ = 3xI X2X3
Giiii
Ta c6: x~ + x~ + x~ = (Xl + X2t + x~ - 3XIX2 (Xl + X2)
= [x, + X2 + X3) [ [x, + X2)2 + X~ - [x, + X2) X3J - 3xI X2 (Xl + X2) Theo dinh 19 Viet: XI + X2 + X3 = 0 ::::::> XI + X2 = -X3
Do d6: x~ + x~ + x~ :::: -3XIX2(-X3) = 3XIX2X3 (dpcm)
Bai 3: Giii sit phirong trlnh: X3 - X2 + ax + b = 0 c6 ba nghiern phan bi~t. Chung minh rhng: a2 + 3b > 0
Giiii
Giii sit phuong trlnh da cho c6 ba nghiern phan biet Xl' X2' X3 .
{XI +X2 +X3 =1
Theo dinh 19 Viet ta c6: . XIX2 + X2X3 + X3XI = a XI.X2.X3 = -b .
79
a2 = (Xl~ + ~Xa + XaX1)2
= (Xlx.J2 + (~Xa)2 + (XaXI)2 + 2Xl ~x.J [x, + ~ + Xa) = (Xtx.J2 + (~Xa)2 + (XaXI)2 - 2b
Theo ba't dAng thtrc Cosi ta duoc:
(X1X2)2 + (X3Xl)2 ~ 2!X1X2!!X3XZ! ~ 2XlX~X3
Tuong nr: (X2X3)2 + (XaXl)2 ~ 2Xl X2X;
(X3Xl)2 + (Xl x2):l ~ 2X;X2X3
Cong v0 theo v6 ba ba't dAng thuc tren ta diroc:
2 2 2
(X1X2) + (X2Xa) + (XaXl) ~ XjX:lXa (Xl + X2 + X3) =-b
Do do: a2 ~ -b - 2b ~ a:l + 3b ~ 0
V1 phirong trlnh dff cho co ba nghiern phan biet nen a2 + 3b > 0 (dpcm)
,-----------------------------------,
Hai 4: Cho phirong trlnh: X3 + Smx" - 3x -- 3m + 2 =: 0 (1)
Dinh m d0 phuong trlnh co ba nghicrn phan biet va hi0u thirc: A = ~ + X; + ~ l __ 2~~_?i,\ trj nM nha'l.
______________ -.J
Giai
Ph irong trtnh (l) ~ (x - 1) [ X2 + (3m + 1) x + 3m - 2] = 0
<c> [ :'=c1( 3m + 1) x+ 3m- 2, 0 (2)
Phuong trlnh (1) co ba nghicm phan hiCt <=> (2) 1..'1) ha I nghi¢m phan hi!)t khac 1.
[tI. > 0
<=:> ' <=:> mt 0
l [(1) of; 0
Khi do: A = xt + x~ + x~ = [x, + X2 + xa)c - 2(XIX2 + x~x;! + X:IX1)
{XI + x , + X, = -3m
Theo dinh ly Viet ta co: ,- , ,. __ ..,
XIX~ + X2X3 + X1X1 -- .,
Do do: A = 9m2 + 6 ~ 6 => min A = 6 da t dl(dc khi III = 0 (loa i) Vh hi6u thirc A khong dat gia tr] nho nhfil.
Hai 5: lim a d0 phirong trtrrh: X3 + x~ + 2ax + -~-: -l)-~'6ha nghicm Xl ,--~~-~ ~'~--J theo thu tif do I~p thanh mOL ca'p s6 nhan,
- ----------- ------
Ghli
Then tinh chfll :'\'p s6 nhal1 va dinh ly Viet ta c6:
80
_ 2 XIX3 - X2
XI +X2 +X3 =-1
_ 2 XIX3 - X2
XI +X3 =-1-x2
XIX3 + (X3 + XI )X2 = 2a XI'X2,X3 =-a
¢:> XIX2 + X2X3 + X3XI = 2a
(1) (2)
{ 3 [a = 0
x --a
TO' (1) vA (2)::) 2 - <=> -Ba3 = -a ee . 1
x =-2a a=±-
2 2../2
[X = 0
+ V"i a = 0 thay vao phU'dng trlnh ta dU'Qc: x3 + x2 = 0 <=>
x =-1
PhU'dng lrlnh chi c6 hai nghi~m nen a = 0 khong thoa man.
+ Vdi a = 2~ thay vao phU'dng trlnh ta dU'~1c:
3 2 1 1 0 (' 1) [:l ( 1) 1 ]
x +x + J2x+2J2= <=> x+ J2 x + 1- J2 x+ 2 =0
1
x=-T2
<=>
x2 + ( 1 - ~) x + .! = 0 ::) phuong trinh nay va nghiem
+ Vdi a = -~ thay vao phU'dng trlnh ta dU'~1C:
2v2 .
x' +x' - ~x- 21 = 0 ¢+- ~)[ x' +( 1+ ~)x+ ~] = 0
1
x2 = J2
~ x, = H -l--i2±Jvi -4)
Ki~m tra lai ta thly: XIX3 = xi
V~y ghi tr] c4n llm'ciia a IA: a = -~ 2v2
81
Bai 6: . Bie't phirong trlnh: ax" + bX2 + ex + 1 = 0 (a *- 0) c6 ba nghiern dimng lA
X1'X2'Xa·
b3 2
Chii . h \, 7 7 7 e
trng min rang: Xl + X2 + X3 ~ - --5 .
81a
Giai
jXl +X2 +X3 =_ b
Theo dinh ly Viet ta c6: a e
XlX2 + X2X3 + X3X1 =a
b3c2 1 ( b)3 e2 1 3 2
Ta c6: ---5 = - -- '-2 = -(Xl + X2 + Xa) (XlX2 + X2Xa + X3Xl)
81a 81 a a 81
Ap dung bit d£ng thirc:
X1X2 + X2X3 + X3Xl ~.!.( Xl + X2 + X3)3 (*)
3
Th~tv~y:bitd£ngthU'c(*)
¢:> 3(XlX2 + X2X3 + X3X1) ~ (Xl + X2 + X3)3
¢:> 2(XlX2 + X2X3 + X3Xl) ~ 2( xi + x; + x;)
<=> (Xl - X2)2 + (X2 - X3)2 + (X3 - Xl)2 ~ 0 luon luon dung.
Do d6: (XlX2 + X2X3 + X3Xl)2 ~.i(Xl + x2 +X3t
<=> 811 (Xl + x2 + xat (Xl~2 + x2xa + Xaxlt ~ ;6 (Xl + X2 + xaf
(do Xl' X2, Xa > 0)
T h' . h 1 ( )7 7 7 7
a c irng nun : 36 Xl + x2 + xa ~ Xl + X2 + Xa
<=> ~(Xl + X2 + X3)8 s (xi + x; + X;)(Xl + X2 + xa) (1) 3
Ta c6: (x. +X2 +X3)2 ~3(x~ +x; + x;)
(theo bit d£ng thtrc Bunhiakopxki)
=> (x. + X2 + X3t ~ 32 (X~ + X~ + xi) ~ 33 (xi + xi + x~) (2)
. [x: + ~ + x: J2 = [R.Jx: + g.jX; + g .fX;]
Lai co: .
~ (xi + X; + ~ ) (Xl + ~ + Xa) (3)
Tlt (1) (2) va (3) suy ra:
82
Bal 7: Cho plnrong trlnh bac ba: X3 - 6X2 + a + a = 0 (1)
Tim a d~ phirong trlnh co ba nghiem phan biet Xl,X2,Xa thoa man:
(Xl - 3)3 + (~ - 3)3 + (Xa - 3)3 = 0 (2)
Giai
Gill sii' (1) c6 ba nghiern phan biet Xl' X2' Xa '
{YI = Xl - 3
E>~ t Y 2 = X2 - 3
Y3 = X3 - 3
Thl Y I - 3; Y 2 - 3; Y 3 - 3 IA nghiern phireng trlnh: (y+3t _6(y+3)2 +a(y+3)+a ='0
~ y3 + 3y2 + (a - 9) y + 4a - 27 = 0
{Yl +Y2 +Y3 =-3
Theo dinh ly Viet ta c6: Y1Y2 + Y2Y3 + Y3Yl = a-9 Yl'Y2'Y3 = -4a+27
(2) ~ y~ + y~ + y~ = 0
~ (YI + Y2 + Y3)3 - 3(YI + Y2 + Y3)(YIY2 + Y2Y3 + Y3YI) + 3YIY2Y3 = 0 ~ -27 + 9 (a - 9) + 3 (27 - 4a) == 0
~a=-9
Thii' lai: Khi a = -9 thl (1) ~ X3 - 6X2 - 9x - 9 = 0 E>~t f(x) = X3 - 6X2 - 9x - 9
. 2 .. [X = 2-J7
f (x) = 3x - 12x - 9; f (x) = 0 ~ r,.;
X = 2 + ,,7
Ta c6: f'(x) = (3x2 -12x - 9)(X -~) -14x -15 = -14x -15 ·33
=> f ( 2 - J7) > 0; f ( 2 + J7) < 0
=> f ( 2 - Ji) .f ( 2 + Ji) < 0
=> phirong trlnh f(x) = 0 luon c6 ba nghiem phan biet.
83
CHUYEN DE 12: cAC D~NG PHUONG TRINH B~C nON CHUAN
I. Cac dang phuong trlnh:
D~NG 1: PHUONG TRINH TRUNG PHUONG
1. Co dang: ax? + bx" + C c- 0 (a 1:- 0)
2. Cac giai:
+ Dc)tt=X2~O
+ Dira vt: phirong trlnh (1.'C hai.
D~NG 2: PHU(1NG TRINH n~c nON CO D~NG (ax +- a)4 + (ax + b)4 = C
1. Cach ghH
1
• D~ t t = ax + 2" (a + b)
• Dira v6 phuong trlnh trung phirong.
2. Ch6 y:
a. NC'u hili loan co tham sf{ m thl ta c6 hai each gid i:
Cdch I: Dung dao ham bang each xct true tic'p ham: . f'(x) = (ax + a)4 + (ax + byt
Ciich 2: Dc)t fin phu r6i dung cong thuc so sanh nghiern.
Tuy nhien, nc'u h9C sinh bie't dao ham thi each 1 sc nhanh han.
b. NC'u bai toan khong co tham s6 thl ta d~t t = ax + ~ (a + b)
c. Ncu d~t t = ax + ~ (a + b) tht ta dung hai each d~ng thirc: (a + b)4 = a" + 4a3b + 6a2b2 + 4ab3 + b4
(a - b)4 = a' - 4a3b + 6a2b2 - 4ab3 + b4
D~NG 3: PHUONG TRINH n~c nON CO D~NG
(x -j a)(x + b)(x + c)(x + d) = k vdi a + b = c + d
1. Cach giai
Buac I: Nhan tung c~p (x + a)(x -+: b); (x + c)(x + d) Buac 2: D~t ftn phu t = x2 + (a + b)x + D
Dira v6 phirong trlnh b~c 2.
84
2. Dang rnd rQng
(AI x2 + BI X + C1 ) (A2X2 + B2x + C2) = k
each 8idi: TIm cac nghicrn cua A1x2 + B1x + C1;A2x2 + B2x + C2 Sau do dira v{j tich.
D~NG 4: PHUONG TRINH B~C BON CO D~NG
(x + a)( x + b)( x + c)( x + d) = k.x2 vdi a.b = c.d
1. Cach giai
Butte i: Nhan tung c~p (x + a)(x + b); (x + c) (x + d)
B '2 Chi h' K K h 2, d t ab uocz: ia ar ve ve cox va ~t = x +-
x
Budc 3: Dira v8 phuong trlnh b~e hai,
2. Chu i. Dang dc}t biet ella phuong trlnh tren la:
a. (AI x2 + BI X + C1 ) (A2X2 + B2x + C2) = k.x2
Ax2 + C + Al x2 + C1 = k
b. Ax2 + BI X + C A2X2 + B2x + C2
(chia tit sCi va m§:u sCi eho x)
D~NG 5: PHUONG TRINH B~C BON CO D~NG
1. ax" + bx3 + cx2 + kbx + k2a = 0
2. ax" ± bx3 + cx2 =F kbx + k2a = 0
1. Cach giai
• Chia hai v€ cua phuong trlnh cho x2 (x *- 0) k
• DAt t = x ±-
. x
D~NG 6: PHUONG TRINH B~C BON CHUA MOT
,q ~, ,,'
BIEU THUC GIONG NHAU
1. Cach giai
• Bi€n d6i d~ phuong trlnh chua mQt bi~u thtrc gi6ng nhau.
• E)~t t = bi~u thirc d6.
• TIm mien gia tri ella t.
• Dira v8 phuong trinh b~c hai thea t.
• D€n day co hai each giai
each J: Dung cong thirc so sanh nghiern.
each 2: Dung khao sat ham sCi va dinh 19 min, max.
85
II. Bai t~p cac dang phU'dng trmh 1. Bal t~p dang 1:
Bai 1: Cho phuong trlnh: X4 - (m + 2)X2 + 4m + 1 = 0 TIm m de phuong trtnh co b6n nghiern phan bier.
(1)
Giai
D~t t = x2 ~ 0
(Ij trd thanh: t2-(m+2)t+4m+1=0
(2)
De (1) co b6n nghiern phan biet thl (2) co hai nghiem phan biet dirong:
{~ > 0 [ 1
-- < m < 0
<=> P>O <=> 4
S> 0 m > 12
(1)
Bai 2: Cho plnrong trlnh: X4 - (m + 2) X2 + 4m + 1 = 0 11m m de phuong trinh c6 b6n nghiern phan biet.
Giai
Cdch. 1: Dung tinh chili cdp slf c~ng va dinh Ij Viet D~t t = x2 ~ O.
(1) tri'1 thanh: e - (3m + l)t + 3m = 0 (2)
De (1) c6 4 nghiern phan biet thl (2) c6 hai nghiem phan biet 0 < tl < t2
J~ > 0
¢:> P>O
[s > 0
{I
m:;t:-
<::> 3
m c- O
(a)
Khi d6 b6n nghiem cua (1) la: -.,Jt;; -.jt;; .jt;; Jt2 B6n nghiern nay l~p thanh ca'p s6 cong
<=> .,Jt; -.jt; =.jt; - (-.jt;) <=> t2 = 9tl.
Theo dinh ly Viet ta du'<;Jc:
1 t1=10(3m+1)
9
t2 = 10 (3m + 1)
t1.t2 = 3m
=> ~(3m + 1)2 = 3m ¢:> 21m2 -82m+ 3 = 0 10
86
<=> m = 3 v m = _.!:_ thmi man (a) 27
Vay cac gia tri cua m dn tlrn la: m= 3 v m. = _!_
.. 27
Cach 2: Thier lIjp bQ nghism
Gill sii' t6n tai s6 a > 0 sao cho bien s6 -3a < -a < a < 3a la b6n nghiern l~p thanh cip'S6 cong ciia phirong trlnh (1)
Khi do: X4 - (3m + 1)x2 + 3m == (x + 3a)(x + a)(x - a)(x - 3a)
<=> X4 _ (3m + 1)x2 + 3m == X4 -10a2x2 + 9a4 D6ng nhit h~ s6 hai ve' ta diroc:
{-(3m + 1) = -10a2 <=>[m = 31
3m = 9a4 m = 27
Vay m = 3 v m = _!_ la cac gia tri dn tim.
. 27 .
Bal 3: Cho phirong trlnh: X4 - 2 (m - 2) x2 + 4 - 3m = 0 TIm m d~ phirong trlnh co hai nghiern phan biet,
(1)
Giai
Cdcb 1: D~t t = X2~ 0
(1)<=> t2 - 2(m - 2)t + 4 - 3m = 0
D~ (1) co hai nghiern phan biet thl (2) co nghiern thod man:
[(2) Co nghiern kep duong
(2) c6 hai nghiem phan biet trai dB:u
(2)
[{Ill - 0
<=> to = m - 2 > 0 tl < 0 < t2
{m2 -m = 0 m>2
<=> <=>
4
m>-
3
4 m>- 3
VA a 'a .. A, r 4
ay c C gt tn cua m can um a: m >"3
Cach2: D~t f'(x) = X4 -2(m-2)x2 +4-3m fl (x) = 4x3 - 4(m - 2)x
fl (x) = 0 <=> [x = 0 2
gtx) = x - m + 2 = 0
Yeu ch bai toan dl1~c thod man:
[(3) vo nghiem va fCO) < 0 .
<=> ~
(3) c6 nghiern kep x = 0 va fCO) < 0
(3)
87
4 <::)m>-
3
Ba14: Cho phirong trlnh: X4 - 2(2m + 1)X2 + m - 2 = 0 (1)
TIm m d6 phirong trlnh c6 ba nghiem phan bi¢t.
GliB
f)~t t = x:! ~ 0
(I) trC1 thanh: t2 - (2m + 1)t + m - 2 = 0 (2)
f)~ (J) c6 ba nghiern pMn bi¢t thl (2) e6 m~t nghi¢m t = 0 vA nghi¢m kia dirong,
• Khi 1 = 0 thl (2) => m = 2
• Khi m = 2 thi (2) <::) t2 _ 5t = 0 <::) [t = 0 t=5>O
V4y m = 2 lA gia tr] cdn tim.
2.Bat tGp d,ng 2
(2)
Bal 5: Oiai cac phuong trlnh:
1. (X+1)4 +(x-3t =32
2. (2x-1)4 +16(x-2t =17
(1)
Gisl
1. Cach 1: f)~ t t = x-I => x = t + 1
(l)trC1thAnh: (t+2)4 +(t-2t =32
<::) 2t4 + 2.6t2.22 + 2.24 = 32 <::) t = 0 <::) x = 1.
V4y phuong trlnh d4 eho c6 nghi~m x = 1.
Cach 2: Dung dao ham:
XCl hAm f'(x) = (x + 1)4 + (x - 3)4
fl (x) = 4 (x + 1)3 + 4 (x - 3t j rex) = 0 ee x = 1 Bang bie'n thien:
x -00
rx
o +
~(x) +00 . __.+oo
----.32~
Theo bang bie'n thien => f(x) ~ 32 'v'x => f(x) = 32 <::) x = 1. V4y phuong tr1nh d4 eho e6 nghi¢m x = 1.
88
2 .. Phirong trlnh (2) ee (2x _1)4 + (2x - 4)4 = 17
E>~ t t = 2x - ~ => 2x = t + ~
2 2
(3)trC1th~nh: (t+%r +(t-%r =17
(3)
Vdi t2 = .!. <::> [t = % => x = ~ .
4 1
t = -- => x = 1
, 2
V4y phirong trlnh da cho c6 hai nghiern: x = 1 v x = ~
Bai 6: Cho phirong trlnh: (x + 2)4 + (x - 4)4 = m TIm m d~ phirong trlnh c6 hai nghiern phdn biet.
Giaii
Xct ham f'(x) = (x+ 2)4 + (X_4)4 vdi X E R
fl (x) = 4 (x; 2)3 + 4 (x - 4)3
f/(x) = 0 <::> 4(x+ 2)3 = -4(x _4)3 <::> x = 1 Bang bi~n thien:
o +
-00
1
+00
+00 +00
f(x) ----..162 _____
Thco hang bie'n thien :::) yeu c~u bai roan dU'(.1c thoa man <=> m > 162
Bai7:ChophU'dngtrlnh: (3x-l)4 +81(x+l)4 =2m (1) j
TIm m d~ hU'dn trlnh c6 hai nghiem x < 0 phil n biC;t. __
Giaii
Xet ham f'(x) = (3x -It + 81(x + It = (3x -It + (3x +- 3f fl (x) = 12(3x _1)3 + 12(3x + 3)3
fl (x) = 0 <::> (3x _1)3 == -(3x + 3)3 ~ x Bang bie'n thien:
3
R9
x -00
1 3
o
f(x)
o +
+00 82
f(x) 32----
Theo bang bien thien ~ phirong trtnh (1) c6 hai nghiern phan biet am <=> 32 < 2m < 82 <=> 16 < m < 41
Bai 7: Cho phuong trlnh: (x -It + 2(4x + 3t = 3m4 (1)
Tim m d~ phuong trlnh c6 nghiern.
Giai
Xet ham so': f'(x) = (x -It + 2( 4x + 3t
fl (x) = 4(x -1)~ + 32(4x + 3)3
fl (x) = 0 <=> (x _1)3 == -( 4x + at <=> x =-! 5
Bang bien thien:
x -00 +00
2
5
f'(x) 0 +
+00 +00
f(x) ~3(H /
Theo bang bien thien ~ (1) c6 nghiern <=> 3.m4 ~ 3.(~r <=> Iml ~ ~
3. Bai t~p dang 3
Bai 9: Giai cac phirong trlnh:
1. (x - 2)(x -1)(x + 3)(x + 4) = 24
2. (X2 + 3x + 2)( x2 + 7x + 12) = 120
(1)
(2)
Giai (1) <=> (x - 2) ( x + 4) (x - 1) ( x + 3) = 24
<=> (X2 + 2x - 8) (X2 + 2x - 3) = 24 D~t t = x2 + 2x + 1 = (x + 1)2 ~ 0
(3) trd thanh: (t - 9)(t - 4) = 24 <=> t2 - 13t + 12 = 0
[t = 1
<=> t = 12
(3)
90
, . (1)2 1 [X = 0
V (11 t = 1 ¢::> X + = ¢::>
x=-2
2 [x = -1 - 2.J3
Vdi t = 12 ¢::> (x + 1) = 12 ¢::>
x = 2.J3-1
V~y phirong trlnh c6 cac nghiem la: x = 0; x = -2; x = -1 ± 2.J3
2. (X2 + 3x + 2)( x2 + 7x + 12) = 120
¢::> (x + 1)( x + 2)( x + 3)( x + 4) = 120 ¢::> (x + l)(x + 4)(x + 2)(x + 3) = 120
¢::> (X2 + 5x + 4)( x2 + 5x + 6) = 120 (4)
2 (5)2 9 9
Dilt t = x + 5x + 4 = x + - - - ~--
. 244
(4) trd thanh: t (t + 2) = 120 ¢::> t2 + 2t -120 = 0
[t = -12 (loai)
¢::> •
t = 10
V di t = 10 ¢::> x2 + 5x + 4 = 10 ¢::> [x = 1 x=-6
V~y phuong trmh c6 cac nghiern la: x = 1 v x = -6
Bai 10: Cho plnrong trlnh: (X2 -1)( x + 3)( x + 5) = m (1)
1. Giiii phirong trlnh khi m = 9.
2. Tim m de phuong trinh c6 bO'n nghiem phan biet xi; X2; X3; X4 thoa man
1 1 1 1
-+-+-+- = -1 (*)
Ghii
Phuong trlnh (1) ¢::> (x -l)(x + l)(x + 3)(x + 5) = m
¢::> (x -l)(x + 5)(x + l)(x + 3) = m
¢::> (x2+4x-5)(x2+4x+3)=m (2)
D~t t = x2 + 4x + 4 = (x + 2)2 ~ 0
(2) tro thanh: (t - 9)(t -1) = m
¢::> t2 -lOt + 9 - m = 0 (3)
1. Khi m = 9 thl (3) ¢::> t2 -lOt = 0 ¢::> [t = 0
t = 10
Vdi t = 0 ¢::> x =-2
91
Vdi t = 10 ~ (x + 2)2 = 10 ~ x = -2 - M v x = M - 2
[X =-2
Vh phirong trlnh e6 cac nghiern la: x = -2 ~ M x=M-2
2. E)6 (1) e6 b6n nghiem phan bi9t thl (3) co hai nghiem phan biet dirong: 0 < t) < t2
~ {~/:oo ~-16<m<9 (a)
8>0
Xl + x2 + X3 + x4 ---1 Khi do: (*) ~
X1X2 X3X4
-4 -4
~ --+--=-1
4 - tl 4 - t2
~ 16-4(t1 +t2)=-4(tl +t2)+t1·t2 ~ tl' t2 = 16 ~ m == -7 thoa man (a) V~y m = -7 la gia tri dn tim.
Bai 11:: Cho phirong trlnh:
(X2 - 3x + 2)( X2 - 9x + 20) = m (II
1. Giai phirong trlnh khi m = 4.
2. TIm m de phirong trlnh co h6n nghiern phan biet XI; X2; X3; X4 sao eho
2 2 2 2 2 30
Xl + x2 + x3 + x4 = X1X2X;;X4-
3. TIm m d6 phirong trlnh co dung hai nghiern.
Giai
(l)~ (x - 1)(x - 2)(x - 4)(x - 5) = m ~ (x2 - 6x + 5)(x2 - 6x + 8) = m E)~t t = x2 - 6x + 9 = (x - 3)2 ~ 0
(2) trd thanh: (t - 4)(t - 1) = m
~ t2 - 5t + 4 - m = 0
(2)
(3)
92
I. Khi m = 4 thi (3) ¢:::> e _ 5t = 0 ¢::> [t = 0,
, t = 5
V di 1 = 0 ¢::> x = 3
Vdi 1 = .5 ¢::> (x - 3)2 = 5¢::> X = 3 ± J5
2.
V~y plurong trinh co de nghicrn la: x = 3 ± J5
D0 (I) co h[)n nuhicrn phan biet thl (3) co hai nghiern 0 < t, < t2
f~ > 0
¢::>lP':'O <:-::>-~<m<4
8>0 4
(a)
ling vrti nghicm l, > 0 th] phirong trlnh: X2 - 6x + 9 = t,
, • • • A , {Xl + x2 = 6
¢:.> x -- ()X + 9 -- l, = 0 co hal nghicrn x.; X2 va
XI,X2 = 9 - tl
Ung vdi nghicrn l~ > 0 thl phirong trlnh x2 - 6x + 9 - t2 = 0 c6 hai nghiern Xl; X4 , {X3 + X4 = 6
va
Xa'X4 = 9 - t2
Thco de: xi + x~ + x~ + x~ = 2Xl X2X3X4 - 30
2 2
¢::> (Xl + X2) - 2Xl X2 + (Xa + X4) -lxax4 = 2XIX2XaX4 - 30
¢::> 36 -- 2 (9 - tl ) + 36 - 2 (9 - t2) = 2 (9 - tl ) (9 - t2) - 30 ¢::> 36+2(tl +t2)=2[81-9(tl +t2)+t1t2]-30
¢::> 36+2.5=2(81-9.5+4-m)-30
25 h ?
¢::> m = - t oa (a)
23
v, 25 I:" ., . ]I ,
ay In = - d gla tn can tim.
. 23 .
3.
D6 (I) co dung hai nghiern thl (3) co nghicrn thoa man:
[(3) c6 nghiern ke p duong
_(3) co hai nghiern trai dau
J~I = 0
. 5 0
¢::> l to = 2" >
P<O
r 9
m = _"_
¢::> 4
m >4
93
Biti 12: Giiii cac phirong trinh:
1. (x-l){x - 2){x - 4)(x - 8) = 10 x2 9
2. (X2 + 3x + 2) (X2 + 9x + 18) = 168x2
(1) (2)
Giai
1. Phuong trlnh (1)
<=> (x -1)(x - 8)(x - 2)(x - 4) = 10 x2 9
<=> (X2 _ 9x + 8)( x2 - 6x + 8) = 10 x2
. 9
<=> (x + : - 9 ) ( x +: - 6 ) = 1~ (x * 0) (*)
E>at t == x + ~ ~ 4./2
. x
(*) trd thanh: (t - 9)(t - 6) = 10 <=> t2 -15t + 476 = 0
9 9
~ [:: :j (nhan)
. ) 7 8 17· 2 17 [X = 3
V di t = - ~ X + - = - <=> x - - x + 8 = 0 <=> 8
3 x 3 3 X=-
3
[X = 14 -Ji24
. 28 8 28 2 28 3
Vdlt~3~x+ x~3~x-3x+8~O~ x~14+f24
2. (x2 +3X+2)(X2 +9x+18)=168x2
<=> (x + l)(x + 2)(x + 3)(x + 6) = 168x2 <=> (X2 + 7x + 6)(X2 + 5x + 5) = 168x2
<=> ( X + : + 7) ( x + : + 5) = 168 (x * 0) (*)
Dat t = x + 6 ~ 2J6
. x
Phuong trinh (*) trd thanh:
94
[t -7
(t + 7)(t + 5) = 168 <=> t2 + 12t -133 = 0 <=> t: -19 (nhan)
ve: t = 7 <=> X2 -7x + 6 = 0 <=> [X = 1 x=6
[ -19 -.J337
Vdi t = -19 <=> X + 6 = -19 <=> X = rn;:;;:;2
X ,,337 -19
x=----
2
V~ h ~ h 6.{ hie 1>' 1 6 -19±.J337
ay p irong trill C cac ng nem a: X = ; X = V X = 2
BM 13: Cho phirong trlnh: (X2 .: X + 1)( X2 - 5x + 1) = mx2 (1) TIm m d~ phimng trlnh c6 nghiern.
Giai
(1) <=> ( X + ~ - 1) ( X + ~ - 5 ) = m (x ,e 0)
f)~t t = X + 1:. ; di~u kien: It I ~ 2
X
(2) trd thanh: (t - 1 )(t - 5) = m <=> e - 6t + 5 = m f)~ (1) c6 nghiern thl (3) c6 nghiern thoa man It I ~ 2
Xet ham f(t) = t2 - 6t + 5
rro = 2t - 6; f(t) = 0 <=> t = 3 Bang bien thien:
t -00 -2
(2)
f(t)
2 3 +00 o +
+00
---'21
-3 +00
-----4 .>
Theo bang bie'n thien => yeu
<=> m ~-4.
BM 14: Cho phirong trlnh: 2 3x + 2 X = m (1)
X - X + 4 2x - 6x + 8
1. Giai phirong trtnh khi m = 1.
2. TIm m d~ phirong trlnh c6 nghiern x > O.
Giai
3 1
(1)<=> 4 + 4 =m . (x,eO)
x+~-l 2(X+~)-6
(2)
95
D~ t t = x + 4 ; di~u kien It I ~ 4 x
3 1
(2) tr<l thanh: -- + -- = m (3) t -1 2t - 6
3 1 [t = 5
1. Khi m = l thl (J)¢.:> -- + -- = 1 ¢.:> 5
t - 1 2t - 6 t = - (loai)
2 .
V di t = 5 ¢.:> x + ! = 5 ¢.:> x2 _ 5x + 4 = 0 ee [X = 1
X x=4
V~y phirong trlnh da: cho co nghiern x = 1 v x = 4.
2. D~t t = X +! VI x > 0 nen t ~ 4 x
D<3 (1) co nghiern x dirong thl (3) co nghiern t ~ 4
xsr ham f'(t) = _3_ + _1_ vdi t ~ 0
t -1 2t - 6
fl (t) = - 3 2 - 2 2 < 0 \it ~ 4
(t - 1) (2t - 6)
Bang hien thien: t 4
+00
3
ret) "2 ____________
o
Theo hang hic'n thicn => yeu du bai roan dU<;1C thod man 3
¢.:>O<m:S;; '2
BailS: Gia] cac phirong trlnh:
1. X4 + 2x3 - 6x2 + 2x + 1 = 0
2. 2X4 - 5x3 + 6x2 - 5x + 2 = 0
(1) (2)
Giai
1. Chia hai vc cua phirong trlnh cho x2 ;1:. 0 du<.1c:
x2 + :2 + 2 ( x + ~ ) - 6 = 0 ¢.:> (x + ~ J + 2 ( x + ~) - 8 ~ 0
D~t t = x + 1:. : ui0u kicn It I ~ 2 x
96
[t - 2
Ta d:u'<;1e: t2 + 2t - 8 = 0 ¢:> t: -4 (nhan)
V ~i t = 2 ¢:> x + .! = 2 ¢:> X = 1
x
Vdi t = -4 ¢:> x +.! = -4 ee x2 + 4x +1 = 0 eex = -2 ±.J3 x
V~y phuong trtnh e6 cac nghiern la: x = -2 ±.J3
2. Chia hai vt eho x ta d:u'<;1e (x *- 0)
2 ( x2 + :2 ) - 5 ( x + ~) + 6 = 0
C> 2( x+ !)' -5( x « !)+2 = 0
[ 1
x+-=2 2
x x -2x+ 1 = 0
¢:> ¢:> .
x + .!:. = ~ [x' - 5x + 2 = 0
x 2
x=l ¢:> x=2 1
X=-
2
BRi 16: Giai cac phtrong trtnh:
1. 9x4 - 30x8 + 15x2 - lOx + 1 = 0
2. 8x4 -10x3 - 6x2 + 5x + 2 = 0
(1) (2)
Gisl
1. Cho hai vt ella phirong trtnh eho x2 :1: 0 ta d:u'<;1e:
9x2 + ~ -10( 3x + ~) + 15 = 0
¢:> ( 3x + ~ r - 10 ( 3x + !) + 9 = 0
[3X + .! = 1 x = 9 - J69
¢:> x¢:> 6
3x + .! = 9 x = 9 + J69 .
x . 6
2. Chia hai vt ella phirong trlnh eho x2 *- 0 ta d:u'<;1e
2(4X2+ :2)-5(2X-!)-6=0
¢:> 2 ( 2x - ~ r - -5 ( 2x - ~) + 2 = 0
97
l2X -.!_ = 2
~ x
1 1 2x-- =-
x 2
[X = 1 - J3 v x = _1 +_J3_3
~ 2 2
x=1-J33 vx=1+J33
8 8
Bai 17: Cho phirong trtnh: X4 - 3mx3 + llx2 - 3mx + 1 = 0 (1)
TIm m d~ phirong trlnh co 4 nghiern phan biet.
Giai Chia hai ve cua (1) cho x2 :;t. 0 ta diroc:
x2 + :2 - 3m( x +~) + 11 = 0
~ (x + ~r - 3m ( x + ~) + 9 = 0 (2)
D~t t = x + .!_ ~ x2 - tx + 1 = 0 x
[t ~-2 Phuong trlnh nay co nghiern ~ ~ ~ 0 ~
t~2
(2)~ t2 - 3mt + 9 = 0 ~ t2 + 9 = 3mt ~ t + ~ = 3m (3)
. t
Xet ham s6 f(t) = t + ~
t
fl (t) = t2 - 9 . fl (t) = 0 ~ t = ±3
t2' .
Bang bien thien:
t -00 -3
+ 0
3
-6
f(x) _007 \ _ 13 2
o
Ung vOi m9t gift IIi t = -2 hay t > 2 thl phirong trlnh t = x +! c6 hai nghiem x phan bi~t x
Do d6 d~ (1) c6 4 nghiern phan biet, thea bang bien thien ta du'<;1C cac gia tri cua m c~n tim Ia:
l· 13 l 13
6 < 3m < - 2 < m < -
2· . 6
13 . ~. 13·
...,.- <: 3m < -6 -- < m < -2
2 . 6
98
Bal 18: Cho phirong trtnh: X4 + ax" + x2 + ax + 1 = 0 (1)
1. Giai phirong trlnh khi a = 4.
2. TIm a d~ phuong trlnh khong co it hen hai nghiern am.
1.
Giiii
VI x = 0 khong Iii nghiern ciia (1) nen chi a hai v€ cua (1) cho x2 ta diroc:
x2 + :2 + a( x + ~) + 1 = 0
D~t t = x + _!'; di~u kien It I ~ 2
x
Phuong trinh tren tro thanh: t2 - 2 + at + 1 = 0 ~ t2 + at - 1 = 0 . (2)
[t = -2-15
Khi a = 4 thl (2) tro thanh: t2 + 4t -1 = 0 ~ r;:
. t = -2 + \15 (loai)
V di t = -2 - 15 => x + _! = -2 - 15
x
[x.=.!.( ~5 + 415 - 2 - 15)
~ x2 + ( 2 + 15) x + 1 = 0 ~ 2
x = -~( 15 + 415 + 2 + 15)
V~lnghi~m c.ua phuong trlnh da cho IA:
x = ~ (-2 - 15 ± ~5 + 415)
2. + Neu t < -2 thl phirong trlnh: x + _! .= t ~ x2 - tx + 1 = 0 IUDn co hai nghiem x
A ,{S = t < -2 < 0 am VI
P=l>O .
+ Neu t > 2 thl phirong trinh: x2 - tx + 1 = 0 co hai nghiern dirong. Nen d~ phirong trlnh (1) co khong it hon hai nghiern am thl phirong trlnh (2) co it nhit mot nghiern t < -2.
t2 -1 Phuong trlnh (2) ~ t2 - 1 = -at ~ -- = -a t
1
fl (t) = 1 + 2" > 0 \it < -2
t
Xet ham s6 f'(t) = t -.!. vdi t <-2 t
Bang bien thien:
t -00
3
f(t) -2"
-00
-2
ret)
+
99
,
.~
Dua vao bang bien thien ta du'<;1c gia tri ciia a c!n tim la:
3 3
-a < - - <=> a > -
2 2
Vay cac gia tri ciia a thoa man d~ bai la: a > ~.
.. 2
fBai 19: Cho phirong trinh: X4 - ax" - (2a + 1) x2 + ax + 1 = 0 (1) L Dinh a d€ phirong trlnh co hai nghiern ldn hon 1.
. Giai
Vi x = 0 khong la nghiern cua (1) nen chia hai v€ clla (1) cho X2 ta du'<;1c:
x2 + _!_ - a (x - .!.) - 2a - 1 = 0
x2 X
1
Dat t = x - -, t E R
. x
Phuong trtnh tren trd thanh: t2 + 2 - at - 2a - 1 = 0 <=> t2 - at - 2a + 1 = 0
+ Vdi a > 1
Xet ham s6:
1 1
t = x - - =::> t' = 1 + "'2 > 0 \Ix > 1
x x
Bang bi€n thien: x 1
t'
+
t 0 +00
Dua vao bang bi€n thien ta suy ra t> 0
1 2
M~t khac: phuong trlnh t = x - - <=> x - tx -1 = 0 luon c6 hai nghiem
x
phan biet: XI < 0 < X2.
Do d6 d€ (1) c6 hai nghiem Idn hon 1 thl (2) c6 hai nghiem phan bi~t dirong,
<=> {~: ~ <=> 2J5 - 4 <a < !
. 2
8>0
VA . a .. k, r 2J5 4 1
ay gi tn cua a can nrn a: - <a<'2'
Bai 20: Chung minh rhng n€u phircng trlnh:
X4 + bx" + cx2 + bx + 1 = 0
co nghiern thi: b2 + (c - 2t > 3
(1)
100
Giai
Gia sti' Xo :;:. 0 la mQt nghiern ciia phirong trlnh thl (1) tirong dirong vdi:
x~ + \ + b (xo + J:...) + c = 0 (2)
Xo Xo
£)~ t: t = Xo + J:... ; di~u kien It I ~ 2
Xo
Khi d6 (2) trd thanh: e + bt + c - 2 = 0 => _t2 = bt + c .;._ 2 Ap dung ba't ddng tlnrc Bunhiakopxki ta duoc:
(_t2)2 =(bt+c-2)2 ~[b2+(c-atJ(e+1) => t2t:1 ~b2+(c_a)2
Ta se chirng minh \;/ It I ~ 2 thl -t- > 3 (3)
t + 1
Th~tvh: (3) <=> t4 >3(t2+1)
4 2 9 21 (2 3)2 21 .. II
<=> t - 3t + - > - <=> t - - - - > 0 vdi mOl t ~ 2
4 4 2 4 .
V~y b2 + (c - 2t > 3
Bai 21: Cho phtrong trlnh: X4 + 2X2 + 2ax + a2 + 2a + 1 = 0 (1)
voi m6i a, goi xa la nghiem be nha't ciia phirong trlnh. Xac dinh a dE Xa nho nha't.
Giai Goi Xa la nghiern ciia phirong trlnh (1) thl:
x! + 2x: + 2axa + a 2 + 2a + 1 = 0
<=> a 2 + 2 [x, + 1) a + x! + 2x: + 1 = 0 (2)
Phuong trlnh nay IA phuong trlnh b~c hai theo in a va phai co nghiem nen: fl.' = [x, + 1)2 - (X! + 2x: + 1) = -x! - x: + 2xa ~ 0
<=> (xa - x!)(x! + Xa + 2) ~ 0 <=>0 ~ Xa ~ 1
=> (Xa )min =: 0 th€ vao (2) ta du'c;1C: a2 + 2a + 1 = 0 <=> a = -1 Vh gia tri a c~n tim la: a = -1.
Chu j: Neu d~ bai c~n nghiern ldn nha't thl tra ICfi Xa = 1 <=> a = -2
6.
BAi 22: Giai phuong trlnh: 3(X2 - x + It - 2(x + It = 5(X3 + 1)
Giai
3 (X2 - X + 1 r - 2 (x + 1 t = 5 (X3 + 1)
101
~ 3 (X2 - X + 1 r - 2 ( x + 1 t = 5 ( x + 1){ x2 - X + 1) (1) + Ne'u x = -1 thl (1) trC1 thanh: 3.9 = 0 volY.
+ Ne'u X;;f:. -1, ~hia hai ve' ciia (1) cho (x + 1){x2 - X + 1) ta du'<;1c:
3. x2 - X + 1 _ 2. x + 1 = 5
x + 1 x2 - X + 1
x2 - X + 1 , x + 1 1
Dill: t = tht =
. X + 1 x2 - X + 1 t
Phuong trlnh tren trC1 thanh: .
3t - ~ = 5 ~ 3t2 - 5t - 2 = 0 ~ [t = _ ~
t t=2
+ V di t = - ~ ~ 3x2 - 2x + 4 = 0 phirong trlnh vo nghiern.
. [x = ~(3 - Jf3)
+ V di t = 2 <=> 2
x =i(3+Jf3)
[X = ~(3- Jf3)
. V~y nghiern plnrong trlnh la: ~ ~ .
x = 2(3+ Jf3)
--=--- =-
x2 + X + 1
3
B" 23 G'" h ' h 3x·
31 : iai p irong trm : -x-=-2---4-x-+-1
2x 8
Giili Di~u kien: x2 - 4x + 1 ;;f:. 0 ~ x ;;f:. 2 ± J3
3 2 8
Phuong trlnh 1 1 = 3
x+--4 x+-+1
x x
1
D~ t: t = x + -; It I ~ 2
x
3 2 8 [t = 5
Phuong trlnh trC1 thanh: -- - -- = - ~ 13
t-4 t+1 3 t=-4
1 [x = i(5.- 51)
+ V di t = 5 ~ x + - = 5 ~ .'
x x = ~ ( 5 + 51)
102
, [x = -13 -'- Jf05
+ Vdi t = _13 <=> x +.! = _13 <=> 8
4 x 4 -13 + .J105
x=----
. 8
Vh nghiern cua phirong trlnh la:
x = '!(5 ± m)· x = -13 ± Jf05
2 ' 8
Bai 24: Giai phirong trlnh: X3 - 3X2 + 3x - 16~ - 9 == 0
Giai
f)~t Y = ~
Phuong trlnh dil cho tr<'1 thanh: y9 - 3y6 + 3y3 -16y - 9 = 0 <=> (y + 1){y2 _ Y _1){y6 + 2y4 - 2y3 -+- 4y2 - 2y + 9) = 0
<=> (y + 1}(y2 - Y _1)[(y3 _1)2 + 2y4 + 3y2 + (y -It + 7] = 0 <=> (y + 1){y2 - y -1) = 0
[y + 1 = 0 .[Y = -1 .
<=> v" _ Y _ 1 = 0 ee y = ~ ( 1 ± Fs)
V di Y == -1 => x = -1
y = ~(1 + Fs) => x = y3 = 2 + Fs
y = ~ ( 1 - Fs) => x = y3 = 2 - Fs
• V~y nghiem dil eho cua phirong trtnh la: x = -1; x = 2 + J5; x = 2 - J5
Bai 25: Gilii phuong trlnh: 11 25 = 1
X2 - (x+5t
Giai
f)i~u kien: {X :;t: 0 x:;t: -5
f)~t: Y = X + 5; y :;t: 0 => X2 = (y - 5t = y2 -lOy + 25
Tbay vao phirong trlnh ta dU'<je: 11 _ 25 = 1
y2 -10y+25 y2
<=> 11y2 _ 25(y2 -lOy + 25) = y2 (y2 -"lOy + 25)
<=> s' -10y3 + 39y2 - 250y + 625 = 0
103
Bai26 G'!' hu1 i h x2+x+1 x2+3x+1 5 (1) : lip eng tr n : X2 + 2x + 1 + x'Ji + 4x + 1 = 6
2625 ( 25)
~ y +-2 -10 y+- +39= 0
. y y
D~t: t = y + 25 ; "i~ukien It I ~ 10
y
=:) t2 = y2 + 625 + 50 =:) y2 + 625 = t2 _ 50
y2 y2
Khi "6 (1) trC1 thanh: t2 -lOt -11 = 0 ~ [t ,.,; -1 [loai] t = 11
(1)
Giaii
a {X2 + 2x + 1 :;t 0 {X:;t -1
Dicu kien: ' ~
X2 + 4x + 1 :;t 0 x :;t -2 ±.J3 '
Vi X = 0 kh6ng IA nghiem eua phirong trinh nen phirong tdnh (1)
1 1
X + - + 1 X + - + 3' 5
X + X =_
116 x+-+2 X:f--+4
X X
D~t t = X + .!. ; "i~u kien It I ~ 2 X
Phinmg trlnh tren trC1 thanh:
t+1 t+3 5
--+--=-
t+2 t+4 6
[ -15 -.J85
t=-~-
~ 7 ,
t = -15;.J85 [loai]
, , -15-.J85 '1 -15'-$5
+ V"1 t = =:) X + - = = a
7 X 7
104
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