\(\Leftrightarrow\left\{{}\begin{matrix}\left(\frac{x}{y+1}\right)^2+\left(\frac{y}{x+1}\right)^2=\frac{1}{2}\\4xy=xy+x+y+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\frac{x}{y+1}\right)^2+\left(\frac{y}{x+1}\right)^2=\frac{1}{2}\\4xy=\left(x+1\right)\left(y+1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\frac{x}{y+1}\right)^2+\left(\frac{y}{x+1}\right)^2=\frac{1}{2}\\\left(\frac{y}{x+1}\right)\left(\frac{x}{y+1}\right)=\frac{1}{4}\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\frac{x}{y+1}=u\\\frac{y}{x+1}=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u^2+v^2=\frac{1}{2}\\uv=\frac{1}{4}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u+v=\pm1\\uv=\frac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left(u;v\right)=\left(\frac{1}{2};\frac{1}{2}\right);\left(-\frac{1}{2};-\frac{1}{2}\right)\)
\(\Rightarrow\left[{}\begin{matrix}\frac{x}{y+1}=\frac{y}{x+1}=\frac{1}{2}\\\frac{x}{y+1}=\frac{y}{x+1}=-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2y=-1\\2x-y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x+2y=-1\\2x+y=-1\end{matrix}\right.\end{matrix}\right.\)