\(\begin{array}{l}A=\left(\frac{1}{\sqrt{x}-3}-\frac{1}{\sqrt{x}+1}\right) \cdot \frac{x-3 \sqrt{x}}{\sqrt{x}} \text { với } x\gt 0 ; x \neq 9 \\ =\frac{\sqrt{x}+1-\sqrt{x}+3}{(\sqrt{x}-3)(\sqrt{x}+1)} \cdot \frac{\sqrt{x}(\sqrt{x}-3)}{\sqrt{x}} \\ =\frac{4}{(\sqrt{x}-3)(\sqrt{x}+1)} \cdot(\sqrt{x}-3)=\frac{4}{\sqrt{x}+1}\end{array}\)