Nhận thấy \(x=0\) không phải nghiệm, chia vế cho vế ta được:
\(\frac{2xy^2+x+2}{2xy-xy^2+2y}=2\Leftrightarrow2xy^2+x+2=4xy-2xy^2+4y\)
\(\Leftrightarrow4xy^2-2\left(2x+2\right)y+x+2=0\)
\(\Delta'=\left(2x+2\right)^2-4x\left(x+2\right)=4\)
\(\Rightarrow\left\{{}\begin{matrix}y=\frac{2x+2+2}{4x}=\frac{x+2}{2x}\\y=\frac{2x+2-2}{4x}=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}xy=\frac{x+2}{2}\\y=\frac{1}{2}\end{matrix}\right.\)
Thay vào pt ban đầu
\(\Rightarrow\left[{}\begin{matrix}2\left(\frac{x+2}{2}\right)^2+x^2+2x=2\\2x^2.\frac{1}{4}+x^2+2x=2\end{matrix}\right.\) \(\Leftrightarrow...\)
