x(x+3)=(3-x)(1+x)
⇔ x² +3x = 3 +2x -x²
⇔ x² +x² +3x -2x -3 = 0
⇔ 2x² +x -3 = 0
⇔ 2x² +3x-2x -3 = 0
⇔ x.(2x +3) - (2x +3) = 0
⇔ (2x +3).(x -1) = 0
⇔ \(\left[ \begin{array}{l}2x+3=0\\x-1=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{-3}{2}\\x=1\end{array} \right.\)
Vậy S= { $\frac{-3}{2}$ ; 1}
