ĐKXĐ: $x∈\Bbb R$
$\sqrt{x^2-2x+1}=x^2-1\\↔\sqrt{(x-1)^2}=x^2-1\\↔|x-1|=x^2-1$
$|x-1|=\begin{cases}x-1\,\,nếu\,\,x-1\ge 0\,\,hay\,\,x\ge 1\\1-x\,\,nếu\,\,x-1<0\,\,hay\,\,x<1\end{cases}$
TH1: $x\ge 1$
$→x-1=x^2-1\\↔x-x^2=0\\↔x(1-x)=0\\↔\left[\begin{array}{1}x=0\\1-x=0\end{array}\right.\\↔\left[\begin{array}{1}x=0(KTM)\\x=1(TM)\end{array}\right.$
TH2: $x<1$
$→1-x=x^2-1\\↔-x^2-x+2=0\\↔x^2+x-2=0\\↔x^2+2x-x-2=0\\↔(x^2+2x)-(x+2)=0\\↔x(x+2)-(x+2)=0\\↔(x-1)(x+2)=0\\↔\left[\begin{array}{1}x-1=0\\x+2=0\end{array}\right.\\↔\left[\begin{array}{1}x=1(KTM)\\x=-2(TM)\end{array}\right.$
Vậy $S=\{1;-2\}$
