\(\begin{array}{l} A = \frac{{\sqrt x - 2}}{{x\sqrt x - 8}} = \frac{{\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {x + 2\sqrt x + 4} \right)}} = \frac{1}{{x + 2\sqrt x + 4}}\\ x \ge 0 \Rightarrow x + 2\sqrt x + 4 \ge 4\\ \Rightarrow A = \frac{1}{{x + 2\sqrt x + 4}} \le \frac{1}{4} \end{array}\)
