ĐKXĐ: \(x\ge9\)
\(A=\frac{\sqrt{x-9}}{5x}=\frac{\sqrt{\frac{x-9}{3}}3-3}{5x}\le\frac{\frac{1}{2}\left(\frac{x-9}{3}+3\right)}{5x}=\frac{\frac{x-9+9}{3}}{10x}=\frac{1}{30}\)
(dấu "=" xảy ra \(\Leftrightarrow\frac{x-9}{3}=3\Leftrightarrow x=18\))
Vậy \(A_{max}=\frac{1}{30}\) (khi và chỉ khi x = 18)