Question

Rút gọn

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

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

 

Answer 1

\(A=\left(\dfrac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right).\dfrac{1}{\sqrt{x}+1}\left(x\ge0;x\ne1\right)\)

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

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

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

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