Question

\(C=(\dfrac{\sqrt x+1}{\sqrt -1}-\dfrac{x+\sqrt x}{x-1})÷\dfrac{1}{\sqrt x+1}+\dfrac{1}{\sqrt x-1}\)

Answer 1

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

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

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

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

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

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