Решебник по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 26

\[\boxed{\mathbf{26.}}\]

\[\textbf{а)}\log_{3}243 \cdot \log_{0,8}{(3x - 5)} > 0\]

\[5\log_{0,8}(3x - 5) > 0\]

\[\log_{0,8}(3x - 5) > 0\]

\[\log_{0,8}{(3x - 5)} > \log_{0,8}{(1)}\]

\[3x - 5 > 0\]

\[3x > 5\]

\[x > \frac{5}{3}.\]

\[M = \left( 1\frac{2}{3}; + \infty \right)\text{.\ }\]

\[3x - 5 < 1\]

\[3x < 6\]

\[x < 2.\]

\[Решение\ неравенства:\]

\[x \in \left( 1\frac{2}{3};2 \right).\]

\[Ответ:\ x \in \left( 1\frac{2}{3};2 \right).\]

\[\textbf{б)}\ \frac{\log_{9}{(5x - 4)}}{\log_{0,4}{0,064}} < 0\]

\[\frac{\log_{9}(5x - 4)}{3} < 0\]

\[\log_{9}{(5x - 4)} < \log_{9}1\]

\[5x - 4 > 0\]

\[5x > 4\]

\[x > 0,8.\]

\[M = (0,8; + \infty).\]

\[5x - 4 < 1\]

\[5x < 5\]

\[x < 1.\]

\[Решение\ неравенства:\]

\[x \in (0,8;1).\]

\[Ответ:\ x \in (0,8;1).\]

\[\textbf{в)}\ \frac{\log_{0,2}{(2 - 5x)}}{\log_{5}625} > 0\]

\[\frac{\log_{0,2}(2 - 5x)}{4} > 0\]

\[\log_{0,2}{(2 - 5x)} > \log_{0,2}1\]

\[2 - 5x > 0\]

\[5x < 2\]

\[x < 0,4.\]

\[M = ( - \infty;0,4).\]

\[2 - 5x < 1\]

\[5x > 1\]

\[x > 0,2.\]

\[Решение\ неравенства:\]

\[x \in (0,2;0,4).\]

\[Ответ:\ x \in (0,2;0,4).\]

\[\textbf{г)}\log_{0,6}{0,216} \cdot \log_{5}{(5 - 2x)} < 0\]

\[3\log_{5}(5 - 2x) < 0\]

\[\log_{5}(5 - 2x) < 0\]

\[\log_{5}{(5 - 2x)} < \log_{5}1\]

\[5 - 2x > 0\]

\[2x < 5\]

\[x < 2,5.\]

\[M = ( - \infty;2,5).\]

\[5 - 2x < 1\]

\[- 2x < - 4\]

\[x > 2.\]

\[Решение\ неравенства:\]

\[x \in (2;2,5).\]

\[Ответ:\ x \in (2;2,5).\]