Question

Giup đi các bro

Answer 1

\(a,ĐK:x>0;x\ne1\\ A=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ b,A>0\Leftrightarrow\sqrt{x}-1>0\left(\sqrt{x}>1\right)\\ \Leftrightarrow x>1\)

Answer 2

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