\(\Leftrightarrow\left\{{}\begin{matrix}x+y+x^2+y^2=8\\\left(x^2+x\right)\left(y^2+y\right)=12\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}u=x^2+x\\v=y^2+y\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u+v=8\\uv=12\end{matrix}\right.\)
Theo Viet đảo, u và v là nghiệm của:
\(t^2-8t+12=0\Rightarrow\left[{}\begin{matrix}t=2\\t=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}u=2\\v=6\end{matrix}\right.\\\left\{{}\begin{matrix}u=6\\v=2\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+x=2\\y^2+y=6\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+x=6\\y^2+y=2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow...\)
