Question

\(P=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{2}{\sqrt{x}-2}\)

Answer 1

\(P=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{\sqrt{x}+2}{x-4}\right)\cdot\dfrac{\sqrt{x}-2}{2}\)

\(=\dfrac{2\sqrt{x}+2}{x-4}\cdot\dfrac{\sqrt{x}-2}{2}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)

Answer 2

\(P=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{2}{\sqrt{x}-2}\\ =\dfrac{\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{2}{\sqrt{x}-2}\\ =\dfrac{2\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{2}{\sqrt{x}-2}\\ =\dfrac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2}{2}\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\) ĐK :\(x\ge0;x\ne4\)

Answer 3

P = \(\left(\dfrac{\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right).\dfrac{\sqrt{x}-2}{2}\)
   = \(\dfrac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{2\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
   = \(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)