\(\Leftrightarrow\left\{{}\begin{matrix}x+y+xy=11\\\left(x+y\right)xy=30\end{matrix}\right.\)
Theo Viet đảo, \(x+y\) và \(xy\) là nghiệm của:
\(t^2-11t+30=0\Rightarrow\left[{}\begin{matrix}t=6\\t=5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=6\\xy=5\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=5\\xy=6\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;5\right);\left(5;1\right);\left(2;3\right);\left(3;2\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+xy=11\\xy\left(x+y\right)=30\end{matrix}\right.\)
Đặt \(x+y=S;xy=P\Rightarrow\left\{{}\begin{matrix}P+S=11\\P.S=30\end{matrix}\right.\)
\(\Rightarrow\frac{30}{S}+S=11\Leftrightarrow30+S^2=11S\Leftrightarrow\left[{}\begin{matrix}S=6\\S=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}P=5\\P=6\end{matrix}\right.\)
Xét \(\left\{{}\begin{matrix}P=6\\S=5\end{matrix}\right.\Rightarrow X^2-5X+6=0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\end{matrix}\right.\)
Làm tương tự vs trường hợp còn lại
