Question

Answer 1

đk : x > 0 ; x khác 1 

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

Answer 2

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

ĐKXĐ: x>=0; x<>1