Đáp án:
\(\left[ \begin{array}{l}
x \ge 3\\
- 3 \le x < 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\sqrt {\dfrac{{{x^2} - 9}}{{x - 1}}} \\
DK:\left\{ \begin{array}{l}
\dfrac{{{x^2} - 9}}{{x - 1}} \ge 0\\
x - 1 \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ne 1\\
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
{x^2} - 9 \ge 0\\
x - 1 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
{x^2} - 9 \le 0\\
x - 1 < 0
\end{array} \right.
\end{array} \right.
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ne 1\\
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 3\\
x \le - 3
\end{array} \right.\\
x > 1
\end{array} \right.\\
\left\{ \begin{array}{l}
- 3 \le x \le 3\\
x < 1
\end{array} \right.
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
x \ge 3\\
- 3 \le x < 1
\end{array} \right.
\end{array}\)
