Bài giải:

$\sqrt{-\frac{x}{x-1}}$ có nghĩa $\Leftrightarrow -\frac{x}{x-1}\ge 0\Leftrightarrow \frac{x}{x-1}\le 0$

TH1: $\left\{ \begin{align}& x\ge 0 \\ & x-1<0\end{align} \right.$

$\Leftrightarrow \left\{ \begin{align}& x\ge 0 \\& x <1 \end{align} \right.\Leftrightarrow 0\le x<1$

TH2: $\left\{ \begin{align}& x\le 0 \\& x-1>0 \\\end{align} \right.$

$\Leftrightarrow \left\{ \begin{align}& x\le 0 \\& x>1 \\\end{align} \right.$(vô lí)

Vậy $0\le x<1$