$B=(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}):\dfrac{\sqrt{x}+1}{(\sqrt{x}-1)^2}$ ĐK: $x>0,x\neq1$
$B=(\dfrac{1}{\sqrt{x}(\sqrt{x}-1)}+\dfrac{1}{\sqrt{x}-1}).\dfrac{(\sqrt{x}-1)^2}{\sqrt{x}+1}$
$B=\dfrac{1+\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}.\dfrac{(\sqrt{x}-1)^2}{\sqrt{x}+1}$
$B=\dfrac{\sqrt{x}-1}{\sqrt{x}}$
Vậy $B=\dfrac{\sqrt{x}-1}{\sqrt{x}}$ với $x>0,x\neq1$.