Bất phương trình mũ \({1 \over {{3^x} + 5}} \le {1...

A

- Hướng dẫn giải

Điều kiện \(x \ne  - 1\)

Ta có: \(\dfrac{1}{{{3^x} + 5}} \le \dfrac{1}{{{3^{x + 1}} - 1}}\)

\(\Leftrightarrow \dfrac{1}{{{3^x} + 5}} - \dfrac{1}{{{3^{x + 1}} - 1}} \le 0\)

\( \Leftrightarrow \dfrac{{{{3.3}^x} - 1 - {3^x} - 5}}{{\left( {{3^x} + 5} \right)\left( {{3^{x + 1}} - 1} \right)}} \le 0\)

\(\Leftrightarrow \dfrac{{{{2.3}^x} - 6}}{{\left( {{3^x} + 5} \right)\left( {{3^{x + 1}} - 1} \right)}} \le 0\)

\( \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}{2.3^x} - 6 \le 0\\{3^{x + 1}} - 1 > 0\end{array} \right.\\\left\{ \begin{array}{l}{2.3^x} - 6 \ge 0\\{3^{x + 1}} - 1 < 0\end{array} \right.\end{array} \right.\)

\(\Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x \le 1\\x >  - 1\end{array} \right.\\\left\{ \begin{array}{l}x \ge 1\\x <  - 1\end{array} \right.\end{array} \right. \Rightarrow x \in \left( { - 1;1} \right]\)