Đáp án:
$S=\{\dfrac{3}{2};-4\}$
Giải thích các bước giải:
$(2x-3)(3x+1)=(2x-3)^2$
$⇔(2x-3)(3x+1)-(2x-3)^2=0$
$⇔(2x-3)[(3x+1)-(2x-3)]=0$
$⇔(2x-3)(3x+1-2x+3)=0$
$⇔(2x-3)(x+4)=0$
\(⇔\left[ \begin{array}{l}2x-3=0\\x+4=0\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=-4\end{array} \right.\)
Vậy $S=\{\dfrac{3}{2};-4\}$