$\begin{array}{l} \dfrac{{x - \sqrt x }}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 1} \right)}} + \dfrac{{\sqrt x + 3}}{{x - 9}}\\ \dfrac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 1} \right)}} + \dfrac{{\sqrt x + 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\ = \dfrac{{\sqrt x }}{{\sqrt x - 3}} + \dfrac{1}{{\sqrt x - 3}} = \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\ b)A = 3 \Leftrightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} = 3\\ \Leftrightarrow \sqrt x + 1 = 3\sqrt x - 9\\ \Leftrightarrow 2\sqrt x = 10 \Leftrightarrow \sqrt x = 5 \Leftrightarrow x = 25\\ c)A = \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} = \dfrac{{\sqrt x - 3 + 4}}{{\sqrt x - 3}} = 1 + \dfrac{4}{{\sqrt x - 3}} \in \mathbb{Z}\\ \Rightarrow 4 \vdots \sqrt x - 3 \Rightarrow \sqrt x - 3 \in U\left( 4 \right) = \left\{ { \pm 1; \pm 2; \pm 4} \right\}\\ \Rightarrow x \in \left\{ {4;16;25;1;49} \right\} \end{array}$
