$\begin{array}{l} A = \dfrac{{\sqrt x + 7}}{{\sqrt x + 2}}\\ A = 1 + \dfrac{5}{{\sqrt x + 2}} > 1\\ A = 1 + \dfrac{5}{{\sqrt x + 2}} \le 1 + \dfrac{5}{2} = \dfrac{7}{2}\\ \Rightarrow 1 < A \le \dfrac{7}{2}\\ A \in Z \Rightarrow \left[ \begin{array}{l} A = 2\\ A = 3 \end{array} \right.\\ + \dfrac{{\sqrt x + 7}}{{\sqrt x + 2}} = 2 \Rightarrow \sqrt x + 7 = 2\sqrt x + 4 \Leftrightarrow \sqrt x = 3 \Leftrightarrow x = 9\\ + \dfrac{{\sqrt x + 7}}{{\sqrt x + 2}} = 3 \Rightarrow \sqrt x + 7 = 3\sqrt x + 6 \Leftrightarrow 2\sqrt x = 1 \Rightarrow x = \dfrac{1}{4} \end{array}$
