x² - 1 = (x+1)(5x + 7)
<=> (x - 1)(x + 1) - (x + 1)(5x + 7) = 0
<=> (x + 1)[(x - 1) - (5x + 7)] = 0
<=> (x + 1)(x - 1 - 5x - 7) = 0
<=> (x + 1)(-4x - 8) = 0
<=> (x + 1)(x + 2) = 0
<=> \(\left[ \begin{array}{l}x+1=0\\x+2=0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=-1\\x=-2\end{array} \right.\)
Vậy S = {-2 ; -1}
