\(2\left(x-1\right)\sqrt{x^2}+2x-1=x^2-2x-1\)
TH1: \(x\ge0\)
\(\Leftrightarrow2\left(x-1\right)x+2x-1=x^2-2x-1\)
\(\Leftrightarrow2x^2-2x+2x-1=x^2-2x-1\)
\(\Leftrightarrow x^2+2x=0\Leftrightarrow x\left(x+2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\left(tm\right)\\x=-2\left(ktm\right)\end{matrix}\right.\)
TH2:\(x< 0\)
\(\Leftrightarrow-2\left(x-1\right)x+2x-1=x^2-2x-1\)
\(\Leftrightarrow-2x^2+2x+2x-1=x^2-2x-1\)
\(\Leftrightarrow3x^2-6x=0\Leftrightarrow3x\left(x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(ktm\right)\end{matrix}\right.\)
Vậy S = {0}
