\(\left\{{}\begin{matrix}2x^2+y^2-3xy=x-y\\-y^2+2x^2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(2x-y-1\right)=0\\2x^2-y^2=1\left(2\right)\end{matrix}\right.\)
Với \(x=y\)
\(pt\left(2\right)\Leftrightarrow2y^2-y^2=1\Leftrightarrow\left(y-1\right)\left(y+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Với \(y=2x-1\)
\(pt\left(2\right)\Leftrightarrow2x^2-\left(2x-1\right)^2=1\Leftrightarrow-2\left(x-1\right)^2=0\)
\(\Leftrightarrow x=1\Leftrightarrow y=1\)
Vậy \(\left[{}\begin{matrix}\left(x;y\right)=\left(-1;-1\right)\\\left(x;y\right)=\left(1;1\right)\end{matrix}\right.\)
