Решебник по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 41

\[\boxed{\mathbf{41.}}\]

\[\textbf{а)}\log_{\frac{1}{11}}\frac{1}{2x - 11} > \log_{\frac{1}{11}}\frac{1}{3x - 20}\]

\[2x - 11 > 0\]

\[2x > 11\]

\[x > 5,5.\]

\[3x - 20 > 0\]

\[3x > 20\]

\[x > \frac{20}{3}\]

\[x > 6\frac{2}{3}.\]

\[M = \left( 6\frac{2}{3}; + \infty \right).\]

\[3x - 20 < 2x - 11\]

\[x < 9.\]

\[Решение\ неравенства:\]

\[x \in \left( 6\frac{2}{3};9 \right).\]

\[Ответ:x \in \left( 6\frac{2}{3};9 \right).\]

\[\textbf{б)}\log_{11}\frac{1}{4x - 7} > \log_{11}\frac{1}{5x - 14}\]

\[4x - 7 > 0\]

\[4x > 7\]

\[x > \frac{7}{4} > 1\frac{3}{4}.\]

\[5x - 14 > 0\]

\[5x > 14\]

\[x > \frac{14}{5} > 2,8.\]

\[M = (2,8; + \infty).\]

\[5x - 14 > 4x - 7\]

\[x > 7.\]

\[Решение\ неравенства:\]

\[x \in (7; + \infty).\]

\[Ответ:x \in (7; + \infty).\]

\[\textbf{в)}\log_{\frac{1}{13}}\frac{1}{4x - 17} > \log_{\frac{1}{13}}\frac{1}{5x - 40}\]

\[4x - 17 > 0\]

\[4x > 17\]

\[x > 4\frac{1}{4}.\]

\[5x - 40 > 0\]

\[5x > 40\]

\[x > 8.\]

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

\[5x - 40 < 4x - 17\]

\[x < 23.\]

\[Решение\ неравенства:\]

\[x \in (8;23).\]

\[Ответ:x \in (8;23).\]

\[\textbf{г)}\log_{13}\frac{1}{3x - 22} > \log_{13}\frac{1}{4x - 13}\]

\[3x - 22 > 0\]

\[3x > 22\]

\[x > 7\frac{1}{3}.\]

\[4x - 13 > 0\]

\[4x > 13\]

\[x > 3\frac{1}{4}.\]

\[M = \left( 7\frac{1}{3}; + \infty \right).\]

\[4x - 13 > 3x - 22\]

\[x > - 9.\]

\[Решение\ неравенства:\]

\[x \in \left( 7\frac{1}{3}; + \infty \right).\]

\[Ответ:x \in \left( 7\frac{1}{3}; + \infty \right).\]