\(6x^4-5x^3-38x^2-5x+6=0\)
Nhận thấy \(x=0\) không phải là nghiệm, chia cả 2 vế cho \(x^2\):
\(6x^2-5x-38-\dfrac{5}{x}+\dfrac{6}{x^2}=0\)
\(\Leftrightarrow6\left(x^2+\dfrac{1}{x^2}\right)-5\left(x+\dfrac{1}{x}\right)-38=0\)
Đặt \(x+\dfrac{1}{x}=a\Rightarrow x^2+\dfrac{1}{x^2}=a^2-2\) pt trở thành:
\(6\left(a^2-2\right)-5a-38=0\)
\(\Leftrightarrow6a^2-5a-50=0\Rightarrow\left[{}\begin{matrix}a=\dfrac{10}{3}\\a=\dfrac{-5}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=\dfrac{10}{3}\\x+\dfrac{1}{x}=\dfrac{-5}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}3x^2-10x+3=0\\2x^2+5x+2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\\x=-2\\x=\dfrac{-1}{2}\end{matrix}\right.\)
