(2x+1)(3x-2) = 4x² -1
⇔ (2x+1)(3x-2)= (2x -1).(2x +1)
⇔ (2x+1)(3x-2) - (2x -1).(2x +1) = 0
⇔ (2x +1).(3x -2 -2x +1) = 0
⇔ (2x +1).(x -1) = 0
⇔ \(\left[ \begin{array}{l}2x+1=0\\x-1=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{-1}{2}\\x=1\end{array} \right.\)
Vậy S= { $\frac{-1}{2}$ ; 1}