Lời giải:
Ta có:
HPT \(\Leftrightarrow \left\{\begin{matrix} 4x^2-2y^2=2\\ xy+x^2=2\end{matrix}\right.\Rightarrow 4x^2-2y^2=xy+x^2\)
\(\Leftrightarrow 3x^2-xy-2y^2=0\)
\(\Leftrightarrow 3x^2-3xy+2xy-2y^2=0\)
\(\Leftrightarrow 3x(x-y)+2y(x-y)=0\Leftrightarrow (3x+2y)(x-y)=0\)
\(\Rightarrow \left[\begin{matrix} x=y\\ x=\frac{-2}{3}y\end{matrix}\right.\)
Nếu $x=y$ thì thay vào PT(1):
\(\Rightarrow 2y^2-y^2=1\Leftrightarrow y^2=1\Leftrightarrow y=\pm 1\Rightarrow x=\pm 1\) (tương ứng)
Nếu $x=\frac{-2}{3}y$, thay vào PT(1):
\(2(-\frac{2}{3}y)^2-y^2=1\Leftrightarrow \frac{-1}{9}y^2=1\Rightarrow y^2< 0\) (vô lý)
Vậy $(x,y)=(\pm 1, \pm 1)$
