Question

rút gọn bt sau: a=\(\left(\dfrac{x-1}{\sqrt{x}-1}+\dfrac{x\sqrt{x}-a}{1-x}\right):\left(\dfrac{\left(\sqrt{x}-1\right)^2+\sqrt{x}}{\sqrt{x}+1}\right)\)

Answer 1

Đoạn $x\sqrt{x}-a$ là sao vậy bạn? Có nhầm lẫn gì không?

Answer 2

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

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

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