Rút gọn căn thức chứa căn bậc hai

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 $x  \geq 0; x \neq 1$)

$= \dfrac{1}{\sqrt{x} - 1} + \dfrac{2}{\left ( \sqrt{x} + 1 \right )\left ( \sqrt{x} - 1 \right )}$

$= \dfrac{\sqrt{x} + 1}{\left ( \sqrt{x} + 1 \right )\left ( \sqrt{x} - 1 \right )} + \dfrac{2}{\left ( \sqrt{x} + 1 \right )\left ( \sqrt{x} - 1 \right )}$

$= \dfrac{\sqrt{x} + 1 + 2}{\left ( \sqrt{x} + 1 \right )\left ( \sqrt{x} - 1 \right )}$

$= \dfrac{\sqrt{x} + 3}{\left ( \sqrt{x} + 1 \right )\left ( \sqrt{x} - 1 \right )}$