Question

Answer 1

 giúp mình vs

 

Answer 2

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

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

Answer 3

\(a,A=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\left(\sqrt{x}-1\right)\left(x>0;x\ne1\right)\\ A=\dfrac{x-1}{\sqrt{x}}\\ b,A< 0\Leftrightarrow x-1< 0\left(\sqrt{x}>0\right)\\ \Leftrightarrow x< 1\Leftrightarrow0< x< 1\)