Trừ vế cho vế: \(3x^2-xy-2y^2=0\)
\(\Leftrightarrow\left(x-y\right)\left(3x+2y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=x\\y=-\frac{3}{2}x\end{matrix}\right.\)
Thay vào pt dưới:
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x^2=2\\x^2-\frac{3}{2}x^2=2\left(vn\right)\end{matrix}\right.\) \(\Leftrightarrow x^2=1\Leftrightarrow...\)