$\dfrac{2}{x+2\sqrt{x}+3}$
$=\dfrac{2}{(\sqrt{x})^2+2.\sqrt{x}.1+1+2}$
$=\dfrac{2}{(\sqrt{x}+1)^2+2}$
Ta thấy: $(\sqrt{x}+1)^2+2≥2$
$→\dfrac{2}{(\sqrt{x}+1)^2+2}≤1$
$→$ Dấu "=" xảy ra khi $\sqrt{x}+1=0$
$→\sqrt{x}=-1$ mà $\sqrt{x}≥0$
$→\sqrt{x}+1=1$
$→x=0$
$→A_{max}=\dfrac{2}{1+2}=\dfrac{2}{3}$