Условие задачи
Решите неравенство:
\(log_5(3x+1)+log_5(\frac{1}{72x^2}+1)\geq log_5(\frac{1}{24x}+1). \)
Решение
\(log_5(3x+1)+log_5(\frac{1}{72x^2}+1)\geq log_5(\frac{1}{24x}+1) < = >\)
\(< = > \left\{\begin{matrix}
3x+1>0;\\
\displaystyle \frac{1}{72x^2}+1>0;\\
\\
\displaystyle \frac{1}{24x}+1>0;\\
\\
log_5((3x+1)\cdot(\displaystyle \frac{1}{72x^2}))+1\geq log_5(\frac{1}{24x}+1)
\end{matrix}\right. < = >\)
\(\left\{\begin{matrix}
x>\displaystyle \frac{1}{3};\\
\\
x\neq 0;\\
\\
\displaystyle \frac{24x+1}{x}>0;\\
\\
(3x+1)\cdot(\displaystyle \frac{1}{72x^2}+1)\geq\frac{1}{24x}+1.
\end{matrix}\right.\)
ОДЗ неравенства:
\(\left[
\begin{gathered}
-\frac{1}{3}
\end{gathered}
\right.\)
Раскроем скобки в последнем неравенстве системы:
\(\displaystyle \frac{1}{24x}+\displaystyle \frac{1}{72x^2}+3x+1\geq \displaystyle \frac{1}{24x}+1;\)
\(\displaystyle \frac{1}{72x^2}+3x \geq 0;\)
\(\displaystyle \frac{1+216x^3}{72x^2}\geq 0.\)
При \(x\neq 0\) получим: \(1+216x^3 \geq 0;\)
\(216x^3 \geq -1;\)
\((6x)^3 \geq (-1)^3;\)
\(6x \geq -1;\)
\(x \geq -\frac{1}{6}.\)
С учетом ОДЗ:
Ответ: \(x\in [-\frac{1}{6};-\frac{1}{24})\cup (0; +\infty).\)
