\(\left\{{}\begin{matrix}x^2-y^2+\dfrac{x}{x-y}=3\left(1\right)\\y^2-2xy-\dfrac{y}{x-y}=-2\left(2\right)\end{matrix}\right.\)
Lấy \(\left(1\right)+\left(2\right)\Rightarrow x^2-2xy+\dfrac{x-y}{x-y}=1\Rightarrow x\left(x-2y\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2y\end{matrix}\right.\)
\(TH_1:x=0\Rightarrow-y^2=3\Rightarrow\) vô lý
\(TH_2:x=2y\Rightarrow4y^2-y^2+\dfrac{2y}{y}=3\Rightarrow3y^2=1\Rightarrow y^2=\dfrac{1}{3}\)
\(\Rightarrow\left[{}\begin{matrix}y=\sqrt{\dfrac{1}{3}}\\y=-\sqrt{\dfrac{1}{3}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\sqrt{\dfrac{1}{3}}\\x=-2\sqrt{\dfrac{1}{3}}\end{matrix}\right.\)
