\(\Leftrightarrow2x^3-2x+x^2-1=4x^2-2x-2\)
\(\Leftrightarrow2x^3-3x^2+1=0\)
\(\Leftrightarrow2x^3-x^2-2x^2+1=0\)
\(\Leftrightarrow2x^2\left(x-1\right)-\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x^2-x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\2x^2-2x+x-1=0\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow2x\left(x-1\right)+\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(S=\left\{1;-\dfrac{1}{2}\right\}\)
\(PT\Leftrightarrow\left(2x+1\right)\left(x-1\right)\left(x+1\right)=2\left(x-1\right)\left(2x+1\right).\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)\left(x+1\right)-2\left(x-1\right)\left(2x+1\right)=0.\\ \Leftrightarrow x+1=0.\\ \Leftrightarrow x=-1.\)
