\(x^2+xy-2y^2=0\Leftrightarrow\left(x-y\right)\left(x+2y\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=y\\x=-2y\end{matrix}\right.\)
Thay xuống pt dưới:
\(\Rightarrow\left[{}\begin{matrix}y^2+3y^2+y=3\\-2y^2+3y^2-2y=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4y^2+y-3=0\\y^2-2y-3=0\end{matrix}\right.\) (bấm máy)