Ta có: \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2\left(xy+1\right)\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x+xy-y=x^2+x-xy-y+2xy+2\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x+xy-y=x^2+x+xy-y+2\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x+xy-y-x^2-x-xy+y-2=0\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2x-2=0\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\\left(y+1\right)^2=\left(y-1\right)\left(y-2\right)-2xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y^2+2y+1=y^2-3y+2+2y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y^2+2y+1-y^2+3y-2-2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\3y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-1\\y=\dfrac{1}{3}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2\left(xy+1\right)\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x^2-x+xy-y=x^2+x-xy-y+2xy+2\\y^2+y-xy-x=y^2-2y+xy-2x-2xy\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}-2x=2\\x+3y=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-1\\-1+3y=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-1\\3y=1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-1\\y=\dfrac{1}{3}\end{matrix}\right.\)
Vậy hpt trên có nghiệm duy nhất (x;y) = (-1; \(\dfrac{1}{3}\))
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\(\left\{{}\begin{matrix}x^2+xy-x-y=x^2+x-xy-y+2xy+2\\y^2-xy-x+y=y^2+xy-2y-2x-2xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2x=2\\x+3y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=3\end{matrix}\right.\)