$\dfrac{2x -1}{x² -4} + \dfrac{x +3}{2 -x} +5 = 0$ `ĐKXĐ:` $x \neq ±2$
$⇔ \dfrac{2x -1 -(x +3).(x +2) +5.(x² -4)}{x² -4} = \dfrac{0}{x² -4}$
$⇒ 2x -1 -(x +3).(x +2) +5.(x² -4) = 0$
$⇔ 2x -1 -x² -5x -6 +5x² -20 = 0$
$⇔ 4x² -3x -27 = 0$
$⇔ 4x.(x -3) +9.(x -3) = 0$
$⇔ (x -3).(4x +9) = 0$
$⇔ \left[ \begin{array}{l}x -3=0\\4x +9=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=3\\x=\dfrac{-9}{4}\end{array} \right.$ $\text {(T/m đkxđ)}$
$\text {Vậy S = {3; $\dfrac{-9}{4}$}}$
