Bài tập nâng cao · Bài 28
Bài tập nâng cao Phương trình, bất phương trình mũ và lôgarit · Bài 28
Giải bất phương trình $\log_{2}\left( \log_{\frac{1}{3}}\frac{3x-7}{x+3} \right)\geq 0$.
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Lời giải
Ta có:
$\begin{array}{l} \log_{2}\left( \log_{\frac{1}{3}}\frac{3x-7}{x+3} \right)\geq 0\Leftrightarrow \left\{ \begin{array}{l} \frac{3x-7}{x+3}>0 \\ \log_{\frac{1}{3}}\frac{3x-7}{x+3}>0 \\ \log_{\frac{1}{3}}\frac{3x-7}{x+3}\geq 1 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} \frac{3x-7}{x+3}>0 \\ \frac{3x-7}{x+3}<1 \\ \frac{3x-7}{x+3}\leq \frac{1}{3} \end{array} \right. \\ \Leftrightarrow \left\{ \begin{array}{l} \frac{3x-7}{x+3}>0 \\ \frac{3x-7}{x+3}\leq \frac{1}{3} \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} \frac{3x-7}{x+3}>0 \\ \frac{8(x-3)}{3(x+3)}\leq 0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} x\in (-\infty ;-3)\cup \left( \frac{7}{3};+\infty \right) \\ x\in (-3;3] \end{array} \right.\Leftrightarrow x\in \left( \frac{7}{3};3 \right]. \end{array}$
$\begin{array}{l} \log_{2}\left( \log_{\frac{1}{3}}\frac{3x-7}{x+3} \right)\geq 0\Leftrightarrow \left\{ \begin{array}{l} \frac{3x-7}{x+3}>0 \\ \log_{\frac{1}{3}}\frac{3x-7}{x+3}>0 \\ \log_{\frac{1}{3}}\frac{3x-7}{x+3}\geq 1 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} \frac{3x-7}{x+3}>0 \\ \frac{3x-7}{x+3}<1 \\ \frac{3x-7}{x+3}\leq \frac{1}{3} \end{array} \right. \\ \Leftrightarrow \left\{ \begin{array}{l} \frac{3x-7}{x+3}>0 \\ \frac{3x-7}{x+3}\leq \frac{1}{3} \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} \frac{3x-7}{x+3}>0 \\ \frac{8(x-3)}{3(x+3)}\leq 0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} x\in (-\infty ;-3)\cup \left( \frac{7}{3};+\infty \right) \\ x\in (-3;3] \end{array} \right.\Leftrightarrow x\in \left( \frac{7}{3};3 \right]. \end{array}$