\(B=(\dfrac{\sqrt x}{\sqrt x+1}-\dfrac{\sqrt x}{x+\sqrt x}):\dfrac{\sqrt x-1}{x-1}\\=(\dfrac{\sqrt x}{\sqrt x+1}-\dfrac{\sqrt x}{\sqrt x(\sqrt x+1)}).\dfrac{x-1}{\sqrt x-1}\\=(\dfrac{\sqrt x}{\sqrt x+1}-\dfrac{1}{\sqrt x+1}).\dfrac{(\sqrt x-1)(\sqrt x+1)}{\sqrt x-1}\\=\dfrac{\sqrt x-1}{\sqrt x+1}.\dfrac{\sqrt x+1}{1}\\=\sqrt x-1\\B<0→\sqrt x-1<0\\↔\sqrt x<1\\↔x<1\\mà\,\,x>0\\→0<x<1\)
Vậy \(0<x<1\)
