\(2y^2-2y^2x-x+x^2-y+xy=1\Leftrightarrow2y^2\left(1-x\right)-x\left(1-x\right)-y\left(1-x\right)=1\Leftrightarrow\left(1-x\right)\left(2y^2-x-y\right)=1\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-x=1\\2y^2-x-y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\2y^2-y-1=0\end{matrix}\right.\) \(\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\) (tmđk) hay \(\left\{{}\begin{matrix}x=0\\y=-\dfrac{1}{2}\end{matrix}\right.\) (ktmđk)
Hoặc \(\left\{{}\begin{matrix}1-x=-1\\2y^2-x-y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y^2-y-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\) (tmđk)
hay \(\left\{{}\begin{matrix}x=2\\y=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy x = 0; y = 1 hoặc x = 2; y = 1
