Answer
$\dfrac{4x}{2x-3}$
Work Step by Step
We are given the complex fraction:
$\dfrac{\dfrac{1}{2x-3}-\dfrac{1}{2x+3}}{\dfrac{1}{2x}-\dfrac{1}{2x+3}}$
Multiply both the numerator and denominator by $2x(2x-3)(2x+3)$ and simplify:
$\dfrac{2x(2x-3)(2x+3)\dfrac{1}{2x-3}-2x(2x-3)(2x+3)\dfrac{1}{2x+3}}{2x(2x-3)(2x+3)\dfrac{1}{2x}-2x(2x-3)(2x+3)\dfrac{1}{2x+3}}$
$=\dfrac{2x(2x+3)-2x(2x-3)}{(2x-3)(2x+3)-2x(2x-3)}$
$=\dfrac{2x(2x+3-2x+3)}{4x^2-9-4x^2+6x}$
$=\dfrac{2x(6)}{6x-9}$
$=\dfrac{12x}{3(2x-3)}$
$=\dfrac{4x}{2x-3}$
