\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-2xy=65\\xy-\left(x+y\right)=17\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\) với \(u^2\ge4v\)
\(\Rightarrow\left\{{}\begin{matrix}u^2-2v=65\\v-u=17\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u^2-2v=65\\v=u+17\end{matrix}\right.\)
Thế pt dưới vào pt trên:
\(\Rightarrow u^2-2\left(u+17\right)=65\)
\(\Leftrightarrow u^2-2u-99=0\Rightarrow\left[{}\begin{matrix}u=11\Rightarrow v=28\\u=-9\Rightarrow v=8\end{matrix}\right.\)
- TH1: \(\left\{{}\begin{matrix}x+y=11\\xy=28\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(7;4\right);\left(4;7\right)\)
TH2: \(\left\{{}\begin{matrix}x+y=-9\\xy=8\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(-1;-8\right);\left(-8;-1\right)\)
