Nhận thấy \(x=0\) không phải nghiệm
\(\Leftrightarrow\left(\frac{x^2+3-3x}{x}\right)\left(\frac{x^2+3-2x}{x}\right)=2\)
\(\Leftrightarrow\left(x+\frac{3}{x}-3\right)\left(x+\frac{3}{x}-2\right)-2=0\)
Đặt \(x+\frac{3}{x}-3=t\)
\(\Rightarrow t\left(t+1\right)-2=0\Leftrightarrow t^2+t-2=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{3}{x}-3=1\\x+\frac{3}{x}-3=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-4x+3=0\\x^2-x+3=0\end{matrix}\right.\)
