$a)A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{x+4}{x-4}\right):\left(\dfrac{2\sqrt{x}+1}{x-2\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\\ =\left(\dfrac{\sqrt{x}(\sqrt{x}-2)}{(\sqrt{x}+2)(\sqrt{x}-2)}-\dfrac{x+4}{x-4}\right):\left(\dfrac{2\sqrt{x}+1}{x-2\sqrt{x}}-\dfrac{\sqrt{x}-2}{(\sqrt{x}-2)\sqrt{x}}\right)\\ =\dfrac{x-2\sqrt{x}-x-4}{x-4}:\dfrac{2\sqrt{x}+1-\sqrt{x}+2}{x-2\sqrt{x}}\\ =\dfrac{-2\sqrt{x}-4}{x-4}:\dfrac{\sqrt{x}+3}{x-2\sqrt{x}}\\ =\dfrac{-2\sqrt{x}-4}{x-4}.\dfrac{x-2\sqrt{x}}{\sqrt{x}+3}\\ =\dfrac{-2\sqrt{x}(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}+2)(\sqrt{x}-2)(\sqrt{x}+3)}\\ =\dfrac{-2\sqrt{x}}{\sqrt{x}+3}\\ b) A > -1\\<=>\dfrac{-2\sqrt{x}}{\sqrt{x}+3} > -1\\ <=>\dfrac{2\sqrt{x}}{\sqrt{x}+3}< 1\\ <=>2\sqrt{x} < \sqrt{x}+3\\ <=>\sqrt{x} < 3\\ <=>0 \le x < 9$
