\(2x^4-x^3-6x^2-x+2=0\\ \Leftrightarrow x^2\left(2x^2-x-6-\dfrac{1}{x}+\dfrac{2}{x^2}\right)=0\\ \Leftrightarrow x^2\left[\left(2x^2+4+\dfrac{2}{x^2}\right)-\left(x+\dfrac{1}{x}\right)-10\right]=0\\ \Leftrightarrow x^2\left[2\left(x^2+2+\dfrac{1}{x^2}\right)-\left(x+\dfrac{1}{x}\right)-10\right]=0\\ \Leftrightarrow x^2\left[2\left(x+\dfrac{1}{x}\right)^2-\left(x+\dfrac{1}{x}\right)-10\right]=0\)
Đặt \(x+\dfrac{1}{x}=y\)
\(\Leftrightarrow x^2\left(2y^2-y-10\right)=0\\ \Leftrightarrow x^2\left(2y^2-5y+4y-10\right)=0\\ \Leftrightarrow x^2\left[\left(2y^2-5y\right)+\left(4y-10\right)\right]=0\\ \Leftrightarrow x^2\left[y\left(2y-5\right)+2\left(2y-5\right)\right]\\ \Leftrightarrow x^2\left(y+2\right)\left(2y-5\right)=0\\ x^2\left(x+\dfrac{1}{x}+2\right)\left(2x+\dfrac{2}{x}-5\right)=0\\ \Leftrightarrow\left(x^2+1+2x\right)\left(2x^2+2-5x\right)=0\\ \Leftrightarrow\left(x^2+2x+1\right)\left(2x^2-x-4x+2\right)=0\\ \Leftrightarrow\left(x+1\right)^2\left[\left(2x^2-x\right)-\left(4x-2\right)\right]=0\\ \Leftrightarrow\left(x+1\right)^2\left[x\left(2x-1\right)-2\left(2x-1\right)\right]=0\\ \Leftrightarrow\left(x+1\right)^2\left(x-2\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)^2=0\\x-2=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x=2\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy tập nghiệm phương trình là \(S=\left\{-1;\dfrac{1}{2};2\right\}\)
