Question

\(\left\{{}\begin{matrix}x^2+y^2=10\\\left(x+1\right)\left(y+1\right)=8\end{matrix}\right.\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-2xy=10\\x+y+xy=7\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\) với \(u^2\ge4v\)

\(\Rightarrow\left\{{}\begin{matrix}u^2-2v=10\\u+v=7\end{matrix}\right.\)

\(\Rightarrow u^2-2\left(7-u\right)=10\)

\(\Leftrightarrow u^2+2u-24=0\) 

\(\Rightarrow\left[{}\begin{matrix}u=4\Rightarrow v=3\\u=-6\Rightarrow v=13\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;3\right);\left(3;1\right)\)