\(\dfrac{\sqrt{-3x}}{x^2-1}=\dfrac{\sqrt{-3x}}{\left(x-1\right)\left(x+1\right)}\)
ĐKXĐ: \(\left\{{}\begin{matrix}-3x\ge0\\\left(x-1\right)\left(x+1\right)\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\x\ne-1\end{matrix}\right.\)