\(1,A=\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{x-4}{\sqrt{x}}\\ =\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}}\\ =\dfrac{2\sqrt{x}}{\sqrt{x}}\\ =2\\ 2,A=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}}{x-4}\\ =\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{x-4}{\sqrt{x}}\\ =\dfrac{2\sqrt{x}}{x-4}\cdot\dfrac{x-4}{\sqrt{x}}\\ =\dfrac{2\sqrt{x}}{\sqrt{x}}\\ =2\\ 3,A=\left(\dfrac{x+3\sqrt{x}}{\sqrt{x}+3}-2\right)\left(\dfrac{x-1}{\sqrt{x}-1}+1\right)\\ =\left[\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\sqrt{x}+3}-2\right]\left[\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+1\right]\\ =\left(\sqrt{x}-2\right)\left(\sqrt{x}+1+1\right)\\ =\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\\ =\left(\sqrt{x}\right)^2-2^2\\ =x-4\)
