Đáp án:
$D = \dfrac{-2}{\sqrt x + 1}$
Giải thích các bước giải:
$\begin{array}{l}D = \dfrac{1}{x - \sqrt x} + \dfrac{1}{x + \sqrt x} + \dfrac{2\sqrt x}{1 - x} \qquad (x >0; \,x \ne 1)\\ \to D = \dfrac{1}{\sqrt x(\sqrt x - 1)} + \dfrac{1}{\sqrt x(\sqrt x + 1)} + \dfrac{2\sqrt x}{(1 - \sqrt x)(1 + \sqrt x)}\\ \to D = \dfrac{\sqrt x + 1}{\sqrt x(\sqrt x + 1)(\sqrt x - 1)} + \dfrac{\sqrt x - 1}{\sqrt x(\sqrt x + 1)(\sqrt x - 1)} - \dfrac{2x}{\sqrt x(\sqrt x + 1)(\sqrt x - 1)}\\ \to D = \dfrac{\sqrt x + 1 + \sqrt x - 1 - 2x}{\sqrt x(\sqrt x + 1)(\sqrt x - 1)}\\ \to D = \dfrac{2\sqrt x - 2x}{\sqrt x(\sqrt x + 1)(\sqrt x - 1)}\\ \to D = \dfrac{2\sqrt x(1 - \sqrt x)}{\sqrt x(\sqrt x + 1)(\sqrt x - 1)}\\ \to D = \dfrac{-2}{\sqrt x + 1}\end{array}$
