Đáp án:
$\rm S=\{-3;\dfrac{8}{3}\}$
Giải thích các bước giải:
$\rm x^2-9+(2x-5)(x+3)=0$
$\rm ⇔(x-3)(x+3)+(2x-5)(x+3)=0$
$\rm ⇔(x+3).(x-3+2x-5)=0$
$\rm ⇔(x+3).(3x-8)=0$
$\rm ⇔$ \(\left[ \begin{array}{l}x+3=0\\3x-8=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-3\\x=\dfrac{8}{3}\end{array} \right.\)
Vậy $\rm S=\{-3;\dfrac{8}{3}\}$