Các bạn ơi xin giúp mình với ????

Đáp án:

$D = \dfrac{-2}{\sqrt x + 1}$

Giải thích các bước giải:

$\begin{array}{l}D = \dfrac{1}{x – \sqrt x} + \dfrac{1}{x + \sqrt x} + \dfrac{2\sqrt x}{1 – x} \qquad (x >0; \,x \ne 1)\\ \to D = \dfrac{1}{\sqrt x(\sqrt x – 1)} + \dfrac{1}{\sqrt x(\sqrt x + 1)} + \dfrac{2\sqrt x}{(1 – \sqrt x)(1 + \sqrt x)}\\ \to D = \dfrac{\sqrt x + 1}{\sqrt x(\sqrt x + 1)(\sqrt x – 1)} + \dfrac{\sqrt x – 1}{\sqrt x(\sqrt x + 1)(\sqrt x – 1)} – \dfrac{2x}{\sqrt x(\sqrt x + 1)(\sqrt x – 1)}\\ \to D = \dfrac{\sqrt x + 1 + \sqrt x – 1 – 2x}{\sqrt x(\sqrt x + 1)(\sqrt x – 1)}\\ \to D = \dfrac{2\sqrt x – 2x}{\sqrt x(\sqrt x + 1)(\sqrt x – 1)}\\ \to D = \dfrac{2\sqrt x(1 – \sqrt x)}{\sqrt x(\sqrt x + 1)(\sqrt x – 1)}\\ \to D = \dfrac{-2}{\sqrt x + 1}\end{array}$