$ĐKXĐ : x \neq -7,-8$
Ta có :
$\dfrac{7}{x+8}+\dfrac{8}{x+7} = \dfrac{x+7}{8} + \dfrac{x+8}{7}$
$\to (x+15).\bigg(\dfrac{1}{x+8}+\dfrac{1}{x+7} - \dfrac{1}{8}-\dfrac{1}{7}\bigg) =0 $
$\to x+15=0$ hoặc $\dfrac{1}{x+8}+\dfrac{1}{x+7} - \dfrac{1}{8}-\dfrac{1}{7}=0$
$\to x=-15$ hoặc $x=0$