\(\sqrt{x^2-2x+1}=x^2-1\\ \Leftrightarrow\sqrt{\left(x-1\right)^2}=\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow\left|x-1\right|=\left(x-1\right)\left(x+1\right)\)(1a)
Ta có 2 TH:
+)TH1:
(1a)\(\Leftrightarrow x-1=\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow\left(x-1\right)-\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(1-x-1\right)=0\\ \Leftrightarrow-x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\) (t/m)
+)TH2:
(1a)\(\Leftrightarrow -(x-1)=\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)+\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)(t/m)
Vậy \(x\in\left\{-2;0;1\right\}\)
