Question

Rut gon:

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

Answer 1

điều kiện xác định : \(x>0;x\ne1\)

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

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

sữa đề chút

Answer 2

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