$\left\{ \begin{array}{l}\frac{x+y}{1+xy}=\frac{1-2y}{2-y} \\ \frac{x-y}{1-xy}=\frac{1-3x}{3-x} \end{array} \right.$
$\left\{ \begin{array}{l}\frac{x+y}{1+xy}=\frac{1-2y}{2-y} \\ \frac{x-y}{1-xy}=\frac{1-3x}{3-x} \end{array} \right.$
$\left\{ \begin{array}{l}\frac{x+y}{1+xy}=\frac{1-2y}{2-y} \\ \frac{x-y}{1-xy}=\frac{1-3x}{3-x} \end{array} \right.$
Giải hệ phương trình$\left\{ \begin{array}{l}\frac{x+y}{1+xy}=\frac{1-2y}{2-y} \\ \frac{x-y}{1-xy}=\frac{1-3x}{3-x} \end{array} \right.$