\(\begin{array}{l}
A = \dfrac{1}{{x - 2\sqrt x + 8}}\,\,\,\left( {x \ge 0} \right)\\
A = \dfrac{1}{{{{\left( {\sqrt x - 1} \right)}^2} + 7}} \le \dfrac{1}{7}\\
{A_{\max }} = \dfrac{1}{7} \Leftrightarrow \sqrt x - 1 = 0 \Leftrightarrow \sqrt x = 1 \Leftrightarrow x = 1\left( {tm} \right)
\end{array}\)