\(\left\{{}\begin{matrix}x+y+xy=5\\\left(x+y\right)^3-3xy\left(x+y\right)=9\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\) với \(u^2\ge4v\) ta được:
\(\left\{{}\begin{matrix}u+v=5\\u^3-3uv=9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}v=5-u\\u^3-3uv=9\end{matrix}\right.\)
\(\Rightarrow u^3-3u\left(5-u\right)=9\)
\(\Leftrightarrow u^3+3u^2-15u-9=0\)
\(\Leftrightarrow\left(u-3\right)\left(u^2+6u+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}u=3\Rightarrow v=2\\u=-3-\sqrt{6}\Rightarrow v=8+\sqrt{6}\left(loại\right)\\u=-3+\sqrt{6}\Rightarrow v=8-\sqrt{6}\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)
