\(A=\frac{1}{\sqrt{x}+3}\left(ĐK:x\ge0\right)\)
Vì \(\sqrt{x}\ge0\forall x\ge0\)
\(\Rightarrow\sqrt{x}+3\ge3\forall x\ge0\)
\(\Rightarrow\frac{1}{\sqrt{x}+3}\le\frac{1}{3}\forall x\ge0\)
\(\Rightarrow A\le\frac{1}{3}\forall x\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow\sqrt{x}=0\)
\(\Leftrightarrow\left(\sqrt{x}\right)^2=0^2\)
\(\Leftrightarrow x=0\left(TMĐK\right)\)
Vậy A đạt GTLN là: \(\frac{1}{3}\Leftrightarrow x=0\)