Question

\((\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}}+1)\div\dfrac{\sqrt{x}}{x+\sqrt{x}}\)

Answer 1

\(\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}}+1\right):\dfrac{\sqrt{x}}{x+\sqrt{x}}\)ĐK : x > 0 

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

Answer 2

Ta có: \(\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}+1\right):\dfrac{\sqrt{x}}{x+\sqrt{x}}\)

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

\(=\dfrac{2x+3\sqrt{x}+1}{\sqrt{x}}\)