Giải thích các bước giải:
a, $A = \dfrac{\sqrt{x}}{\sqrt{x} - 1} - \dfrac{1}{x - \sqrt{x}}$ (ĐK $x > 0$)
$= \dfrac{\sqrt{x}}{\sqrt{x} - 1} - \dfrac{1}{\sqrt{x}\left ( \sqrt{x} - 1 \right )}$
$= \dfrac{x}{\sqrt{x}\left ( \sqrt{x} - 1 \right )} - \dfrac{1}{\sqrt{x}\left ( \sqrt{x} - 1 \right )}$
$= \dfrac{x - 1}{\sqrt{x}\left ( \sqrt{x} - 1 \right )}$
$= \dfrac{\left ( \sqrt{x} + 1 \right )\left ( \sqrt{x} - 1 \right )}{\sqrt{x}\left ( \sqrt{x} - 1 \right )}$
$= \dfrac{\sqrt{x} + 1}{\sqrt{x}}$
b, $B = \dfrac{1}{\sqrt{x} - 1} + \dfrac{2}{x - 1}$ (ĐK $x \geq 0; x \neq 1$)
$= \dfrac{1}{\sqrt{x} - 1} + \dfrac{2}{\left ( \sqrt{x} + 1 \right )\left ( \sqrt{x} - 1 \right )}$
$= \dfrac{\sqrt{x} + 1}{\left ( \sqrt{x} + 1 \right )\left ( \sqrt{x} - 1 \right )} + \dfrac{2}{\left ( \sqrt{x} + 1 \right )\left ( \sqrt{x} - 1 \right )}$
$= \dfrac{\sqrt{x} + 1 + 2}{\left ( \sqrt{x} + 1 \right )\left ( \sqrt{x} - 1 \right )}$
$= \dfrac{\sqrt{x} + 3}{\left ( \sqrt{x} + 1 \right )\left ( \sqrt{x} - 1 \right )}$
