\(x^2-2x+1=3x\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)^2-3x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-1-3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy..............................................
\(x^2-2x+1=3x\left(x-1\right)\)
\(< =>\left(x-1\right)^2-3x\left(x-1\right)=0\)
\(< =>\left(x-1\right)\left(x-1-3x\right)=0\)
\(< =>\left(x-1\right)\left(-2x-1\right)=0\)
\(< =>\left[{}\begin{matrix}x-1=0\\-2x-1=0\end{matrix}\right.\)
\(< =>\left[{}\begin{matrix}x=1\\x=\dfrac{-1}{2}\end{matrix}\right.\)
S=\(\left\{1,\dfrac{-1}{2}\right\}\)