Tập nghiệm của bất phương trình \(\frac{{\log \left( {{x^2} - 9} \right)}}{{\log \left( {3 - x} \right)}} \le 1\) là:

* Hướng dẫn giải

Điều kiện: \(\left\{ \begin{array}{l}
{x^2} - 9 > 0\\
3 - x > 0\\
\log \left( {3 - x} \right) \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x > 3\\
x <  - 3
\end{array} \right.\\
x < 3\\
3 - x \ne 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x <  - 3\\
x \ne 2
\end{array} \right. \Leftrightarrow x <  - 3\)

\(\begin{array}{l}
\frac{{\log \left( {{x^2} - 9} \right)}}{{\log \left( {3 - x} \right)}} \le 1 \Leftrightarrow \frac{{\log \left( {{x^2} - 9} \right) - \log \left( {3 - x} \right)}}{{\log \left( {3 - x} \right)}} \le 0 \Leftrightarrow \frac{{\log \frac{{{x^2} - 9}}{{3 - x}}}}{{\log \left( {3 - x} \right)}} \le 0\\
 \Leftrightarrow \frac{{\log \left[ { - \left( {x + 3} \right)} \right]}}{{\log \left( {3 - x} \right)}} \le 0 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\log \left[ { - \left( {x + 3} \right)} \right] \ge 0\\
\log \left( {3 - x} \right) < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
\log \left[ { - \left( {x + 3} \right)} \right] \le 0\\
\log \left( {3 - x} \right) > 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
 - x - 3 \ge 1\\
3 - x < 1
\end{array} \right.\\
\left\{ \begin{array}{l}
 - x - 3 \le 1\\
3 - x > 1
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \le  - 4\\
x > 4
\end{array} \right.\\
\left\{ \begin{array}{l}
x \ge  - 4\\
x < 2
\end{array} \right.
\end{array} \right. \Leftrightarrow  - 4 \le x < 2 \Rightarrow  - 4 \le x <  - 3
\end{array}\)