Question

Answer 1

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

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

Answer 2

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

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

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