Answer
Solution set = $\emptyset$.
(no solution)
Work Step by Step
First, we exclude those values of x that yield a zero in any of the denominators.
$x\not\in\{-2,+2 \}\qquad (*)$
The last denominator $x^{2}-4$ is factored as $(x-2)(x+2)$
Multiply the equation with the LCD=$(x-2)(x+2)$
$ 1(x+2)+1(x-2)=4\qquad$ ... simplify (distribute)
$x+2+x-2=4$
$2x=4$
$ x=2 \qquad$ ...Checking (*), this is not a valid solution
Solution set = $\emptyset$.
(no solution)
