ĐK: $x\ne 5,x\ge 0$
$B\,=\left(\dfrac{15-\sqrt x}{x-25}+\dfrac{2}{\sqrt x+5}\right):\dfrac{\sqrt x+1}{\sqrt x-5}\\\quad =\left(\dfrac{15-\sqrt x}{(\sqrt x-5)(\sqrt x+5)}+\dfrac{2}{\sqrt x+5}\right):\dfrac{\sqrt x+1}{\sqrt x-5}\\\quad =\left(\dfrac{15-\sqrt x}{(\sqrt x-5)(\sqrt x+5)}+\dfrac{2(\sqrt x-5)}{(\sqrt x-5)(\sqrt x+5)}\right).\dfrac{\sqrt x-5}{\sqrt x+1}\\\quad =\dfrac{(15-\sqrt x)+2(\sqrt x-5)}{(\sqrt x-5)(\sqrt x+5)}.\dfrac{\sqrt x-5}{\sqrt x+1}\\\quad =\dfrac{15-\sqrt x+2\sqrt x-10}{\sqrt x+5}.\dfrac{1}{\sqrt x+1}\\\quad =\dfrac{\sqrt x+5}{\sqrt x+5}.\dfrac{1}{\sqrt x+1}\\\quad =\dfrac{1}{\sqrt x+1}$
Vậy $B=\dfrac{1}{\sqrt x+1}$ với $x\ge 0,x\ne 5$
Ta có: $\sqrt x\ge 0$
$↔\sqrt x+1\ge 1\\↔\sqrt x+1>0\\→\dfrac{1}{\sqrt x+1}>0\\↔B>0\\→B=|B|$
Vậy $B=|B|$
