$\begin{array}{l}A = \left(\dfrac{1}{\sqrt x - 1} + \dfrac{1}{\sqrt x + 1} \right):\dfrac{1}{\sqrt x - 1}\qquad (x \geq 0;\, x \ne 1)\\ a)\quad A = \dfrac{1}{\sqrt x - 1} :\dfrac{1}{\sqrt x - 1} + \dfrac{1}{\sqrt x +1}:\dfrac{1}{\sqrt x -1}\\ \to A = 1 +\dfrac{\sqrt x -1}{\sqrt x +1}\\ \to A = \dfrac{2\sqrt x}{\sqrt x+1}\\ b)\quad Xét\,\,A -2\\ = \dfrac{2\sqrt x}{\sqrt x + 1} - 2\\ = \dfrac{2\sqrt x - 2(\sqrt x + 1)}{\sqrt x + 1}\\ = \dfrac{-2}{\sqrt x +1}\\ \text{Ta có:}\\ \sqrt x \geq 0\\ \to \sqrt x + 1 \geq 1 >0\\ \to -\dfrac{2}{\sqrt x + 1} <0\\ \to A - 2 <0\\ \to A < 2\\ c)\quad x = 9 +4\sqrt2\\ \to x = (2\sqrt2 + 1)^2\\ \to \sqrt x = 2\sqrt2 + 1\\ \to A = \dfrac{2.(2\sqrt2 +1)}{2\sqrt2 +1 +1}\\ \to A = \dfrac{2\sqrt2 +1}{\sqrt2 +1}\\ \to A = \dfrac{(\sqrt2 +1)(2 +\sqrt2 +1)}{\sqrt2 +1}\\ \to A = 3 + \sqrt2\\ d)\quad A = \dfrac{2\sqrt x}{\sqrt x + 1} \in \Bbb Z\\ \to A = 2 -\dfrac{2}{\sqrt x + 1} \in \Bbb Z\\ \to \dfrac{2}{\sqrt x+1} \in \Bbb Z\\ \to \sqrt x + 1 \in Ư(2)=\{-2;-1;1;2\}\\ mà \,\,\sqrt x + 1 \geq 1\\ nên\,\,\sqrt x + 1 = \{1;2\}\\ \to \sqrt x = \{0;1\}\\ \to x = \{0;1\}\\ mà\, \,x \ne 1\\ nên\,\,x=0 \end{array}$