ĐKXĐ : \(\left\{{}\begin{matrix}x+1\ne0\\y-1\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne-1\\y\ne1\end{matrix}\right.\)
Ta có : \(\left\{{}\begin{matrix}\frac{x^2}{x+1}+\frac{y^2}{y-1}=4\\\frac{x+2}{x+1}+\frac{y-2}{y-1}=y-x\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{x^2+2x+1-2x-1}{x+1}+\frac{y^2-2x+1+2x-1}{y-1}=4\\\frac{x+1+1}{x+1}+\frac{y-1-1}{y-1}=y-x\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{-2x-1}{x+1}+x+1+\frac{2y-1}{y-1}+y-1=4\\\frac{1}{x+1}+1-\frac{1}{y-1}+1=y-x\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{-2x-2+1}{x+1}+x+1+\frac{2y-2+1}{y-1}+y-1=4\\\frac{1}{x+1}+1-\frac{1}{y-1}+1=y-x\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}-2+\frac{1}{x+1}+x+1+2+\frac{1}{y-1}+y-1=4\\\frac{1}{x+1}+1-\frac{1}{y-1}+1=y-x\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{1}{x+1}+\frac{1}{y-1}=4-x-y\\\frac{1}{x+1}-\frac{1}{y-1}=y-x-2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{1}{x+1}+\frac{1}{y-1}=4-x-y\\\frac{2}{x+1}=2-2x\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{1}{x+1}+\frac{1}{y-1}=4-x-y\\\frac{1}{x+1}=1-x\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{1}{x+1}+\frac{1}{y-1}=4-x-y\\1-x^2=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{1}{x+1}+\frac{1}{y-1}=4-x-y\\-x^2=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{1}{1}+\frac{1}{y-1}=4-y\\x=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{1}{y-1}=3-y\\x=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}3y-3-y^2+y=1\\x=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y^2-4y+4=0\\x=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left(y-2\right)^2=0\\x=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=2\\x=0\end{matrix}\right.\) ( TM )
Vậy phương trình trên có nghiệm là ( x;y ) = ( 0;2 )
