(6x-3)².(1+2x) = 0
⇔ \(\left[ \begin{array}{l}6x-3=0\\1+2x=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}6x=3\\2x=-1\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{1}{2}\\x=\frac{-1}{2}\end{array} \right.\)
Vậy S= { $\frac{1}{2}$ ; $\frac{-1}{2}$ }