Question

Giải hệ phương trình sau

\(\left\{{}\begin{matrix}x^2+2y^2-3=0\\x^2+xy-2=0\end{matrix}\right.\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}2x^2+4y^2-6=0\\3x^2+3xy-6=0\end{matrix}\right.\)

\(\Rightarrow x^2+3xy-4y^2=0\Rightarrow\left(x-y\right)\left(x+4y\right)=0\Rightarrow\left[{}\begin{matrix}x=y\\x=-4y\end{matrix}\right.\)

- Với \(x=y\Rightarrow3x^2-3=0\Rightarrow x=y=\pm1\)

- Với \(x=-4y\Rightarrow18y^2-3=0\Rightarrow\left[{}\begin{matrix}y=\frac{1}{\sqrt{6}}\Rightarrow x=\frac{-4}{\sqrt{6}}\\y=\frac{-1}{\sqrt{6}}\Rightarrow x=\frac{4}{\sqrt{6}}\end{matrix}\right.\)