Đáp án:
$\frac{5}{x-3}$ + $\frac{4}{x+3}$ + $\frac{x-5}{9-x²}$ = 0
ĐKXĐ : x ≠ ±3
⇔ $\frac{5(x+3)}{(x-3)(x+3)}$ + $\frac{4(x-3)}{(x-3)(x+3)}$ - $\frac{x-5}{(x-3)(x+3)}$ = 0
⇒ 5x + 15 + 4x - 12 - x + 5 = 0
⇔ 5x + 4x - x = -5 + 12 - 15
⇔ 8x = -8
⇔ x = -1
Vậy S = { -1}