ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\frac{x}{y+1}\right)^2+\left(\frac{y}{x+1}\right)^2=\frac{1}{2}\\4xy=xy+x+y+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\frac{x}{y+1}\right)^2+\left(\frac{y}{x+1}\right)^2=\frac{1}{2}\\4xy=\left(x+1\right)\left(y+1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\frac{x}{y+1}\right)^2+\left(\frac{y}{x+1}\right)^2=\frac{1}{2}\\\left(\frac{x}{y+1}\right)\left(\frac{y}{x+1}\right)=\frac{1}{4}\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\frac{x}{y+1}=a\\\frac{y}{x+1}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2+b^2=\frac{1}{2}\\ab=\frac{1}{4}\end{matrix}\right.\) \(\Rightarrow a=b=\pm\frac{1}{2}\)
TH1: \(a=b=\frac{1}{2}\Rightarrow\left\{{}\begin{matrix}\frac{x}{y+1}=\frac{1}{2}\\\frac{y}{x+1}=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=y+1\\2y=x+1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x-y=1\\x-2y=-1\end{matrix}\right.\) (bấm máy)
TH2: \(a=b=-\frac{1}{2}\) tương tự
