ĐKXĐ: \(x;y\ne0\)
\(x^2y+2y+x=4xy\Leftrightarrow x+\frac{2}{x}+\frac{1}{y}=4\)
Đặt \(\left\{{}\begin{matrix}\frac{1}{x}=a\\\frac{1}{y}=b\end{matrix}\right.\) \(\Rightarrow\frac{x}{y}=\frac{b}{a}\) ( hệ trở thành:
\(\left\{{}\begin{matrix}\frac{1}{a}+2a+b=4\\a^2+ab+\frac{b}{a}=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a+\frac{1}{a}+a+b=4\\a^2+1+b\left(a+\frac{1}{a}\right)=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+\frac{1}{a}+a+b=4\\a\left(a+\frac{1}{a}\right)+b\left(a+\frac{1}{a}\right)=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a+\frac{1}{a}+a+b=4\\\left(a+\frac{1}{a}\right)\left(a+b\right)=4\end{matrix}\right.\)
\(\Rightarrow\) Theo Viet đảo, \(a+\frac{1}{a}\) và \(a+b\) là nghiệm của:
\(t^2-4t+4=0\Rightarrow t=2\Rightarrow\left\{{}\begin{matrix}a+\frac{1}{a}=2\\a+b=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
