Answer
This equation has no solution
Work Step by Step
$\dfrac{1}{x+2}=\dfrac{4}{x^{2}-4}-\dfrac{1}{x-2}$
Factor the denominator of the second fraction:
$\dfrac{1}{x+2}=\dfrac{4}{(x-2)(x+2)}-\dfrac{1}{x-2}$
Multiply the whole equation by $(x-2)(x+2)$:
$(x-2)(x+2)\Big[\dfrac{1}{x+2}=\dfrac{4}{(x-2)(x+2)}-\dfrac{1}{x-2}\Big]$
$x-2=4-x-2$
Take all terms to the left side of the equation and simplify it by combining like terms:
$x-2-4+x+2=0$
$2x-4=0$
Solve for $x$:
$2x=4$
$x=2$
Substituting $x=2$ in the original equation makes both denominator on the right side $0$, this means that this equation has no solution.
