Answer
$x=4$
Work Step by Step
The values of $x$ that satisfy the equation $
\dfrac{4}{x-2}-\dfrac{x}{x+2}=\dfrac{16}{x^2-4}
$ are
\begin{array}{l}
\dfrac{4}{x-2}-\dfrac{x}{x+2}=\dfrac{16}{(x+2)(x-2)}
\\\\
(x+2)(x-2)\left( \dfrac{4}{x-2}-\dfrac{x}{x+2} \right)=\left( \dfrac{16}{(x+2)(x-2)} \right)(x+2)(x-2)
\\\\
(x+2)(4)-(x-2)(x)=16(1)
\\\\
4x+8-(x^2-2x)=16
\\\\
4x+8-x^2+2x=16
\\\\
-x^2+6x-8=0
\\\\
x^2-6x+8=0
\\\\
(x-4)(x-2)=0
\\\\
x=\{ 2,4 \}
.\end{array}
Upon checking, only $
x=4
$ satisfies the given equation.
