$|x(x+1)|=|3-x|$
$⇔$\(\left[ \begin{array}{l}x(x+1)=3-x\\x(x+1)=x-3\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x^2+x=3-x\\x^2+x=x-3\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x^2+2x+1=4\\x^2=-3⇒LOẠI (do x^2 \geq 0)\end{array} \right.\)
$⇔(x+1)^2=4$
$⇔$\(\left[ \begin{array}{l}x+1=2\\x+1=-2\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=1\\x=-3\end{array} \right.\)
