$A=\dfrac{2\sqrt{x}+3}{\sqrt{x}+1}$
$=\dfrac{2\sqrt{x}+2}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}+1}$
$=2+\dfrac{1}{\sqrt{x}+1}$
Vì `\sqrt{x}>=0⇔\sqrt{x}+1>=1`
$⇒\dfrac{1}{\sqrt{x}+1}<=1$
`->`$2+\dfrac{1}{\sqrt{x}+1}<=3$
Dấu "=" xảy ra khi `\sqrt{x}+1=1⇔x=0`
Vậy `A_(Max)=3` khi `x=0`