对于任意$$\left( x,y \right)\in M$$，均有$$\textbar x\textbar+\textbar y\textbar\leqslant 1$$，故$$\begin{cases}\textbar x\textbar\leqslant 1    \textbar y\textbar\leqslant 1    \end{cases}$$，故有$$\begin{cases}{{x}^{2}}\leqslant \textbar x\textbar{}    {{y}^{2}}\leqslant \textbar y\textbar{}    \end{cases}\Rightarrow {{x}^{2}}+{{y}^{2}}\leqslant \textbar x\textbar+\textbar y\textbar$$，  故$$\left( x,y \right)\in N$$，所以$$M\subseteq N$$；  又因为$$\left( 1,1 \right)\in N$$，$$\left( 1,1 \right)
otin M$$，  故$$M\subsetneqq N$$，选$$\text{B}$$． 
已知集合$$M=\left { \left( x,y \right)\textbar\textbar x\textbar+\textbar y\textbar\leqslant 1 \right }$$，集合$$N=\left { \left( x,y \right)\textbar{{x}^{2}}+{{y}^{2}}\leqslant \textbar x\textbar+\textbar y\textbar{} \right }$$，则．
a. $$M=N$$ 
b. $$M\subseteq N$$ 
c. $$N\subseteq M$$ 
d. 前三个都不正确 