Giải thích các bước giải:
$B=(\dfrac{2x+1}{x\sqrt{x}-1}+\dfrac{1}{1-\sqrt{x}}):(1-\dfrac{x-2}{x+\sqrt{x}+1})$
$\to B=(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}):(1-\dfrac{x-2}{x+\sqrt{x}+1})$
$\to B=(\dfrac{2x+1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}-\dfrac{x+\sqrt{x}+1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}):(1-\dfrac{x-2}{x+\sqrt{x}+1})$
$\to B=(\dfrac{2x+1-x-\sqrt{x}-1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}):(1-\dfrac{x-2}{x+\sqrt{x}+1})$
$\to B=\dfrac{x-\sqrt{x}}{(\sqrt{x}-1)(x+\sqrt{x}+1)}:(1-\dfrac{x-2}{x+\sqrt{x}+1})$
$\to B=\dfrac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}:(1-\dfrac{x-2}{x+\sqrt{x}+1})$
$\to B=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}:\dfrac{x+\sqrt{x}+1-x+2}{x+\sqrt{x}+1}$
$\to B=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}:\dfrac{\sqrt{x}+3}{x+\sqrt{x}+1}$
$\to B=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}.\dfrac{x+\sqrt{x}+1}{\sqrt{x}+3}$
$\to B=\dfrac{\sqrt{x}}{\sqrt{x}+3}$
