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