Answer
$x=-5$
Work Step by Step
$\dfrac{x}{x-2}+\dfrac{1}{x+2}=\dfrac{8}{x^{2}-4}$
Factor the denominator of the fraction on the right side:
$\dfrac{x}{x-2}+\dfrac{1}{x+2}=\dfrac{8}{(x-2)(x+2)}$
Multiply the whole equation by $(x-2)(x+2)$:
$(x-2)(x+2)\Big[\dfrac{x}{x-2}+\dfrac{1}{x+2}=\dfrac{8}{(x-2)(x+2)}\Big]$
$x(x+2)+(x-2)=8$
$x^{2}+2x+x-2=8$
Take $8$ to the left side:
$x^{2}+2x+x-2-8=0$
$x^{2}+3x-10=0$
Solve by factoring:
$(x+5)(x-2)=0$
Set both factor equal to $0$ and solve each individual equation for $x$:
$x+5=0$
$x=-5$
$x-2=0$
$x=2$
The original equation is undefined for $x=2$, so the only solution for this equation is $x=-5$