由题设得：$$\begin{cases}x-1=0    {{x}^{2}}+{{y}^{2}}+2x-4=0   \end{cases}$$或$$\begin{cases}x-1\textgreater0    x+y-1=0    {{x}^{2}}+{{y}^{2}}+2x-4=0   \end{cases}$$，  有三组解$$\left( 1,1 \right)$$，$$\left( 1,-1 \right)$$，$$\left( \frac{\sqrt{6}}{2},1-\frac{\sqrt{6}}{2} \right)$$．  故选$$\text{C}$$． 
方程组$$\begin{cases}\left( x+y-1 \right)\sqrt{x-1}=0    {{x}^{2}}+{{y}^{2}}+2x-4=0   \end{cases}$$的解的个数为．
(A) $$1$$ 
(B) $$2$$ 
(C) $$3$$ 
(D) $$4$$ 