\(A=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{2}{\sqrt{x}+2}\left(x\ge0;x\ne4\right)\)
\(=\left[\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right]\cdot\dfrac{\sqrt{x}+2}{2}\)
\(=\dfrac{2\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+2}{2}\)
\(=\dfrac{2\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
\(Toru\)
\(A=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{2}{\sqrt{x}+2}\left(dk:x\ge0,x\ne4\right)\\ =\dfrac{\sqrt{x}+\sqrt{x}+2}{x-4}.\dfrac{\sqrt{x}+2}{2}\\ =\dfrac{2\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{2}\\ =\dfrac{2\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}-2\right)}\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
