$A>0(x\ge 0)$ hay $\dfrac{\sqrt x-1}{\sqrt x+3}>0$

$↔\sqrt x-1>0$ (do $\sqrt x+3>0\forall x\ge 0$)

Xét hiệu $B-3(x\ge 0,x\ne 1)$:

$=\dfrac{x-\sqrt x+1}{\sqrt x-1}-3\\=\dfrac{x-\sqrt x+1-3\sqrt x+3}{\sqrt x-1}\\=\dfrac{x-4\sqrt x+4}{\sqrt x-1}\\=\dfrac{(\sqrt x-2)^2}{\sqrt x-1}$

Ta có: $\begin{cases}(\sqrt x-2)^2\ge 0\forall x\ge 0,x\ne 1\\\sqrt x-1>0(cmt)\end{cases}$

$→\dfrac{(\sqrt x-2)^2}{\sqrt x-1}\ge 0$ hay $B-3\ge 0$

$↔B\ge 3$

Vậy $B\ge 3$

$#Coder$