Đáp án:
c. \( - 3 \le a \le 3\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:\left\{ \begin{array}{l}
- 3x \ge 0\\
{x^2} - 1 \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \le 0\\
x \ne \pm 1
\end{array} \right.\\
b.DK:{a^2} - 9 \ge 0\\
\to \left( {a - 3} \right)\left( {a + 3} \right) \ge 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
a - 3 \ge 0\\
a + 3 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
a - 3 \le 0\\
a + 3 \le 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
a \ge 3\\
a \ge - 3
\end{array} \right.\\
\left\{ \begin{array}{l}
a \le 3\\
a \le - 3
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
a \ge 3\\
a \le - 3
\end{array} \right.\\
c.DK:9 - {a^2} \ge 0\\
\to \left( {3 - a} \right)\left( {3 + a} \right) \ge 0\\
\to - 3 \le a \le 3
\end{array}\)
