\(\left\{{}\begin{matrix}x^2y+5=-3\\xy^2-7=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2y=-8\\xy^2=8\end{matrix}\right.\)
Nhân theo vế của 2 pt ta có:
\(x^3y^3=-64\Rightarrow\left(xy\right)^3=\left(-4\right)^3\Rightarrow xy=-4\)
\(\Rightarrow x^2y=-8\Leftrightarrow x\cdot xy=-8\Leftrightarrow x\cdot\left(-4\right)=-8\Rightarrow x=2\)
Thì \(xy=-4\Leftrightarrow y\cdot2=-4\Rightarrow y=-2\)
Vậy \(x=2;y=-2\)