ĐKXĐ: \(x\ge0\)
\(\dfrac{x}{\sqrt{x}-1}>0\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\\sqrt{x}-1>0\end{matrix}\right.\)
\(\Leftrightarrow x>1\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có: \(\dfrac{x}{\sqrt{x}-1}>0\)
mà x>0
nên \(\sqrt{x}-1>0\)
hay x>1