Bài giải:

Thay a = -3 vào phương trình ta có:

$\frac{x+(-3)}{-3-x}+\frac{x-(-3)}{-3+x}=\frac{-3[3.(-3)+1]}{{{(-3)}^{2}}-{{x}^{2}}}$

ĐKXĐ:  x $\ne $3; x $\ne $- 3

$\Leftrightarrow \frac{x-3}{-3-x}+\frac{x+3}{x-3}=\frac{24}{9-{{x}^{2}}}$

$\Leftrightarrow \frac{3-x}{x+3}+\frac{x+3}{x-3}=\frac{24}{(3-x)(3+x)}$

$\Leftrightarrow $$\frac{(3-x)(x-3)}{(x+3)(x-3)}+\frac{{{(x+3)}^{2}}}{(x-3)(x+3)}=\frac{-24}{(x+3)(x-3)}$ 

$\Leftrightarrow $-x² + 6x – 9 + x² + 6x + 9 = -24

$\Leftrightarrow $12x = -24

$\Leftrightarrow $x = -2   (thỏa mãn)

Vậy x = -2