Thay \(x=0\) vào pt thấy không phải là nghiệm, chia 2 vế cho \(x^2\):
\(\left(\frac{2x^2-3x+1}{x}\right)\left(\frac{2x^2+5x+1}{x}\right)=9\)
\(\Leftrightarrow\left(2x+\frac{1}{x}-3\right)\left(2x+\frac{1}{x}+5\right)-9=0\)
Đặt \(2x+\frac{1}{x}-3=a\Rightarrow2x+\frac{1}{x}+5=a+8\) pt trở thành:
\(a\left(a+8\right)-9=0\Leftrightarrow a^2+8a-9=0\)
\(\Rightarrow\left[{}\begin{matrix}a=1\\a=-9\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2x+\frac{1}{x}-3=1\\2x+\frac{1}{x}-3=-9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-4x+1=0\\2x^2+6x+1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{2\pm\sqrt{2}}{2}\\x=\frac{-3\pm\sqrt{7}}{2}\end{matrix}\right.\)
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