\(\left\{{}\begin{matrix}x^2+xy+y=19\left(1\right)\\x-xy+y=-1\left(2\right)\end{matrix}\right.\)
Từ (2) <=> xy=x+y+1 thế vào (1) ta được
\(\left(x^2+y^2+2xy\right)-xy=19\) <=> \(\left(x+y\right)^2-\left(x+y\right)-20=0\) Đặt x+y=t ta đc
\(t^2-t-20=0\)\(\) <=> \(\left[{}\begin{matrix}t+4=0\\t-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=-4\\t=5\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x+y=-4\\x+y=5\end{matrix}\right.\) thế vào (2) ta đc
\(\left[{}\begin{matrix}x+y=-4\\xy=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\pm\sqrt{7}\\x=-2\mp\sqrt{7}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x+y=5\\x-y=6\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\y=3\end{matrix}\right.\) hoặc \(\left[{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)
