Question

giúp m với m cảm ơn

Answer 1

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

Answer 2

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

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

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