Nếu $x≥0$
$⇒x=x$
$⇒x^2-x-6=0$
$⇔x^2-3x+2x-6=0$
$⇔x(x-3)+2(x-3)=0$
$⇔(x-3)(x+2)=0$
$⇔$\(\left[ \begin{array}{l}x-3=0\\x+2=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=3(t/m)\\x=-2(loại)\end{array} \right.\)
Nếu $x<0$
$⇒x=-x$
$⇒x^2+x-6=0$
$⇔x^2+3x-2x-6=0$
$⇔x(x+3)-2(xx+3)=0$
$⇔(x+3)(x-2)=0$
$⇔$\(\left[ \begin{array}{l}x+3=0\\x-2=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=-3(t/m)\\x=2(loại)\end{array} \right.\)
Vậy $S=${$±3$}
