$B=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}$
$ B=$ $1-\dfrac{2}{\sqrt{x}+3}$
Ta có: $\sqrt{x}≥0$
$⇒{\sqrt{x}+3}≥3$
$⇒\dfrac{2}{\sqrt{x}+3}≤\dfrac{2}{3}$
$⇒1-\dfrac{1}{\sqrt{x}+3}≤\dfrac{1}{3}$
$\to \dfrac{1}{B}$ $\geq$ $3$
Vậy GTNN của $ \dfrac{1}{B}=3$ khi `x=0`