Question

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

Rút gọn theo lớp 9 hộ mik nhé

Answer 1

ĐKXĐ : \(0\le x\ne1\)

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

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

Vậy ... 

Answer 2

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