Nhận thấy \(x=0\) ko phải nghiệm, chia 2 vế cho \(x^2\):
\(2x^2-5x+6-\frac{5}{x}+\frac{2}{x^2}=0\)
\(\Leftrightarrow2\left(x^2+\frac{1}{x^2}\right)-5\left(x+\frac{1}{x}\right)+6=0\)
Đặt \(x+\frac{1}{x}=a\Rightarrow x^2+\frac{1}{x^2}=a^2-2\)
\(2\left(a^2-2\right)-5a+6=0\)
\(\Leftrightarrow2a^2-5a+2=0\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{x}=2\\x+\frac{1}{x}=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-2x+1=0\\2x^2-x+2=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow x=1\)
