\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+3x\right)\left(y^2+2y\right)=-6\\x^2+3x+y^2+2y=1\end{matrix}\right.\)
Theo Viet đảo, \(x^2+3x\) và \(y^2+2y\) là nghiệm của:
\(t^2-t-6=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-2\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x^2+3x=3\\y^2+2y=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2+3x-3=0\\\left(y+1\right)^2+1=0\end{matrix}\right.\) (vô nghiệm)
Th2: \(\left\{{}\begin{matrix}x^2+3x=-2\\y^2+2y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2+3x+2=0\\y^2+2y-3=0\end{matrix}\right.\) \(\Leftrightarrow...\)