Đáp án:
$\dfrac{3}{x^2+3x} + \dfrac{x-9}{9-x^2}$
$\text{ĐKXĐ : x $\neq$ 0 ; x $\neq$ ± 3}$
$ = \dfrac{3(x-3)}{x(x+3)(x-3)} + \dfrac{-x(x-9)}{x(x-3)(x+3)}$
$ = \dfrac{3(x-3)-x(x-9)}{x(x+3)(x-3)}$
$ = \dfrac{3x-9 -x^2 +9x}{x(x+3)(x-3)}$
$=\dfrac{-x^2+12x-9}{x(x+3)(x-3)}$
$ = \dfrac{x^2-12x+9}{-x^3 +9x}$