Ta có
$\left( \dfrac{x-3}{x} - \dfrac{x}{x-3} + \dfrac{9}{x(x-3)} \right) : 2x - \dfrac{2}{x}$
$= \left( \dfrac{(x-3)^2}{x(x-3)} - \dfrac{x^2}{x(x-3)} + \dfrac{9}{x(x-3)} \right):2x - \dfrac{2}{x}$
$= \dfrac{x^2-6x+9 - x^2 + 9}{x(x-3).2x} - \dfrac{2}{x}$
$= \dfrac{-6x+18}{2x^2(x-3)} - \dfrac{2}{x}$
$= \dfrac{-6}{2x^2} - \dfrac{2}{x}$
$= \dfrac{-6}{2x^2} - \dfrac{2.2x}{2x^2}$
$= \dfrac{-6-4x}{2x^2} = \dfrac{-3-2x}{x^2}$
