$\frac{7}{x+1}+\frac{x+4}{2x-2}=\frac{3x^2-38}{x^2-1}$ $Đkxđ:x\neq±1$
$⇒14x-14+x^2+x+4x+4=6x^2-76$
$⇔5x^2-19x-66=0$
$⇔(5x^2-30x)+(11x-66)=0$
$⇔(x-6)(5x+11)=0$
$⇔\left[ \begin{array}{l}x-6=0\\5x+11=0\end{array} \right.⇔\left[ \begin{array}{l}x=6(tm)\\x=\frac{-11}{5}(tm)\end{array} \right.$
Vậy $S=\{6;\frac{-11}{5}\}$.