$\sqrt{x^2-2x+1}=x^2-1$
$\Rightarrow \sqrt{(x-1)^2}=(x-1)(x+1)$
$\Rightarrow|x-1|=(x-1)(x+1)$
TH1: $x-1\ge0\Leftrightarrow x\ge1$
$\Rightarrow x-1=(x-1)(x+1)$
$\Rightarrow (x-1)(x+1)-(x-1)=0$
$\Rightarrow (x-1)x=0$
$\Rightarrow \left[ \begin{array}{l} x-1=0 \\ x=0 \end{array} \right .\Rightarrow \left[ \begin{array}{l} x=1 (tm)\\ x=0\text{ (loại)}\end{array} \right .$
TH2: $x-1<0\Rightarrow x<1$
$\Rightarrow -(x-1)=(x-1)(x+1)$
$\Rightarrow (x-1)(x+1)+(x-1)=0$
$\Rightarrow (x-1)(x+2)=0$
$\Rightarrow \left[ \begin{array}{l} x-1=0 \\ x+2=0 \end{array} \right .\Rightarrow \left[ \begin{array}{l} x=1 \text{ (loại)}\\ x=-2(tm)\end{array} \right .$
Vậy phương trình có nghiệm $x=\{1;-2\}$.
