(x-1)²-1+x²=(1-x)(x+3)
⇔(x-1)²+(x+1)(x-1)=-(x-1)(x+3)
⇔(x-1)²+(x+1)(x-1)+(x-1)(x+3)=0
⇔(x-1)(x-1+x+1+x+3)=0
⇔(x-1)(3x+3)=0
⇔3(x-1)(x+1)=0
⇔\(\left[ \begin{array}{l}x-1=0\\x+1=0\end{array} \right.\)⇔\(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
Vậy S={±1}.