Question

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

Answer 1

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

⇔\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\\xy=-x-3\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\\\left[{}\begin{matrix}x=-3\\y=0\end{matrix}\right.\end{matrix}\right.\)