ĐKXĐ: \(x;y\ne-1\)
\(3xy=x+y+1\Leftrightarrow4xy=xy+x+y+1\)
\(\Leftrightarrow4xy=\left(x+1\right)\left(y+1\right)\Leftrightarrow\frac{y}{\left(x+1\right)}.\frac{x}{\left(y+1\right)}=\frac{1}{4}\)
Đặt \(\left\{{}\begin{matrix}\frac{x}{y+1}=a\\\frac{y}{x+1}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2+b^2=\frac{1}{2}\\a.b=\frac{1}{4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)^2-2ab=\frac{1}{2}\\ab=\frac{1}{4}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=\pm1\\ab=\frac{1}{4}\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}a+b=1\\ab=\frac{1}{4}\end{matrix}\right.\) \(\Rightarrow\) theo Viet đảo, \(a;b\) là nghiệm của pt:
\(t^2-t+\frac{1}{4}=0\Rightarrow t=\frac{1}{2}\Rightarrow a=b=\frac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{y+1}=\frac{1}{2}\\\frac{y}{x+1}=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow x=y=1\)
TH2: \(\left\{{}\begin{matrix}a+b=-1\\ab=\frac{1}{4}\end{matrix}\right.\) \(\Rightarrow\) a; b là nghiệm pt:
\(t^2+t+\frac{1}{4}=0\Rightarrow t=\frac{-1}{2}\Rightarrow a=b=\frac{-1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{y+1}=\frac{-1}{2}\\\frac{y}{x+1}=\frac{-1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x+y+1=0\\x+2y+1=0\end{matrix}\right.\) \(\Rightarrow x=y=\frac{-1}{3}\)
