Ta có
$\left( 2x + 1 - \dfrac{1}{1-2x} \right) : \left( 2x - \dfrac{4x^2}{2x-1} \right) = \dfrac{(2x+1)(1-2x) - 1}{1-2x} : \dfrac{2x(2x-1) - 4x^2}{2x-1}$
$= \dfrac{1 - 4x^2 - 1}{1-2x} : \dfrac{4x^2 - 2x - 4x^2}{2x-1}$
$= \dfrac{-4x^2}{1-2x} . \dfrac{2x-1}{-2x}$
$= \dfrac{-2x}{1-2x} .(2x-1)$
$= 2x$