\( |x|=\begin{cases}x\,\,nếu\,\,x\ge 0\\-x\,\,nếu\,\,x<0\end{cases}\\TH1:x\ge 0 →x^2-x-6=0\\↔x^2-3x+2x-6=0\\↔x(x-3)+2(x-3)=0\\↔(x+2)(x-3)=0\\↔\left[\begin{array}{1}x+2=0\\x-3=0\end{array}\right.\\↔\left[\begin{array}{1}x=-2(KTM)\\x=3(TM)\end{array}\right.\\TH2:x<0→x^2+x-6=0\\↔x^2+3x-2x-6=0\\↔x(x+3)-2(x+3)=0\\↔(x-2)(x+3)=0\\↔\left[\begin{array}{1}x-2=0\\x+3=0\end{array}\right.\\↔\left[\begin{array}{1}x=2(KTM)\\x=-3(TM)\end{array}\right.\)
Vậy \(S=\{-3;3\}\)
