1 câu trả lời
Đáp án: $x < - 1\,hoặc\,0 \le x < 1$
Giải thích các bước giải:
$\begin{array}{l}
\sqrt {\dfrac{{ - 3x}}{{{x^2} - 1}}} \\
Dkxd:\dfrac{{ - 3x}}{{{x^2} - 1}} \ge 0\\
\Leftrightarrow \dfrac{{3x}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} \le 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \le 0\\
\left( {x - 1} \right)\left( {x + 1} \right) > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x \ge 0\\
\left( {x - 1} \right)\left( {x + 1} \right) < 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \le 0\\
\left[ \begin{array}{l}
x > 1\\
x < - 1
\end{array} \right.
\end{array} \right.\\
\left\{ \begin{array}{l}
x \ge 0\\
- 1 < x < 1
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x < - 1\\
0 \le x < 1
\end{array} \right.\\
Vay\,x < - 1\,hoặc\,0 \le x < 1
\end{array}$