Answer
$\frac{16}{x-4}$.
Work Step by Step
The given expression is
$=\frac{2x}{x-4}+\frac{64}{x^2-16}-\frac{2x}{x+4}$
First denominator $=(x-4)$.
Second denominator $=x^2-16$.
Factor $=x^2-4^2$
Use the algebraic identity $a^2-b^2=(a+b)(a-b)$.
$=(x+4)(x-4)$
Third denominator $=(x+4)$
Substitute all factors into the given expression.
$=\frac{2x}{(x-4)}+\frac{64}{(x+4)(x-4)}-\frac{2x}{(x+4)}$
The LCM of both denominators is $=(x+4)(x-4)$.
$=\frac{2x}{(x-4)}\times \frac{x+4}{x+4}+\frac{64}{(x+4)(x-4)}-\frac{2x}{(x+4)}\times \frac{x-4}{x-4}$
Simplify.
$=\frac{2x(x+4)}{(x-4)(x+4)}+\frac{64}{(x+4)(x-4)}-\frac{2x(x-4)}{(x+4)(x-4)}$
$=\frac{2x(x+4)+64-2x(x-4)}{(x-4)(x+4)}$
$=\frac{2x^2+8x+64-2x^2+8x}{(x-4)(x+4)}$
Simplify.
$=\frac{16x+64}{(x-4)(x+4)}$
Factor the numerator as shown below.
$=16x+64$
$=16(x+4)$
Substitute into the fraction.
$=\frac{16(x+4)}{(x-4)(x+4)}$
Cancel common terms.
$=\frac{16}{x-4}$.
