Решебник по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 33

\[\boxed{\mathbf{33.}}\]

\[\textbf{а)}\ 4\log_{x}3 > \log_{3}x - 3\]

\[M = (0;1) \cup (1; + \infty).\]

\[\frac{4}{\log_{3}x} > \log_{3}x - 3\]

\[4 < \log_{3}^{2}{x - 3}\log_{3}x\]

\[\log_{3}^{2}x - 3\log_{3}x - 4 > 0\]

\[\log_{3}x = t:\]

\[t^{2} - 3t - 4 > 0\]

\[t_{1} + t_{2} = 3;\ \ t_{1} \cdot t_{2} = - 4\]

\[t_{1} = - 1;\ \ \ t_{2} = 4;\]

\[(t + 1)(t - 4) > 0\]

\[t < - 1;\ \ t > 4.\]

\[0 < \log_{3}x < 1:\]

\[\log_{3}x < - 1\]

\[\log_{3}x < \log_{3}\frac{1}{3}\]

\[x < \frac{1}{3}.\]

\[0 < \log_{3}x < 4:\]

\[\log_{3}1 < \log_{3}x < \log_{3}81\]

\[1 < x < 81.\]

\[Решение\ неравенства:\]

\[x \in \left( 0;\frac{1}{3} \right) \cup (1;81).\]

\[Ответ:\ x \in \left( 0;\frac{1}{3} \right) \cup (1;81).\]

\[\textbf{б)}\ 2\log_{x}5 > \log_{5}x - 1\]

\[M = (0;1) \cup (1; + \infty).\]

\[\frac{2}{\log_{5}x} > \log_{5}x - 1\]

\[2 < \log_{5}^{2}x - \log_{5}x\]

\[\log_{5}^{2}x - \log_{5}x - 2 > 0\]

\[t = \log_{5}x:\]

\[t^{2} - t - 2 > 0\]

\[t_{1} + t_{2} = 1;\ \ t_{1} \cdot t_{2} = - 2\]

\[t_{1} = - 1;\ \ t_{2} = 2;\]

\[(t + 1)(t - 2) > 0\]

\[t < - 1;\ \ t > 2.\]

\[0 < \log_{5}x < 1:\]

\[\log_{5}x < - 1\]

\[\log_{5}x < \log_{5}\frac{1}{5}\]

\[x < \frac{1}{5}.\]

\[0 < \log_{5}x < 2\]

\[\log_{5}1 < \log_{5}x < \log_{5}25\]

\[1 < x < 5.\]

\[Решение\ неравенства:\]

\[x \in \left( 0;\frac{1}{5} \right) \cup (1;25).\]

\[Ответ:\ x \in \left( 0;\frac{1}{5} \right) \cup (1;25).\]