Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1613

\[\boxed{\mathbf{1613}\mathbf{.}}\]

\[1)\log_{\frac{1}{2}}(1 + x) - \sqrt{x^{2} - 4} \leq 0\]

\[\log_{\frac{1}{2}}(1 + x) \leq \sqrt{x^{2} - 4}\ \ \ \ \ | \bullet ( - 1)\]

\[\log_{2}(1 + x) \geq - \sqrt{x^{2} - 4}\]

\[при\ любом\ x.\]

\[Имеет\ смысл\ при:\]

\[1 + x > 0\]

\[x > - 1.\]

\[x^{2} - 4 \geq 0\]

\[x^{2} \geq 4\]

\[x \leq - 2\ \ и\ \ x \geq 2.\]

\[Ответ:\ \ x \geq 2.\]

\[2)\ \frac{1}{\log_{5}(3 - 2x)} - \frac{1}{4 - \log_{5}(3 - 2x)} < 0\]

\[y = \log_{5}(3 - 2x):\]

\[\frac{1}{y} - \frac{1}{4 - y} < 0\]

\[\frac{4 - y - y}{y(4 - y)} < 0\]

\[2 \bullet \frac{2 - y}{y(4 - y)} < 0\]

\[y \bullet (2 - y) \bullet (4 - y) < 0\]

\[y \bullet (y - 2) \bullet (y - 4) < 0\]

\[y < 0\ \ и\ \ 2 < y < 4.\]

\[1)\ \log_{5}(3 - 2x) < 0\]

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

\[3 - 2x < 1\]

\[2x > 2\]

\[x > 1.\]

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

\[\log_{5}(3 - 2x) > \log_{5}5^{2}\]

\[3 - 2x > 25\]

\[2x < - 22\]

\[x < - 11.\]

\[3)\ \log_{5}(3 - 2x) < 4\]

\[\log_{5}(3 - 2x) < \log_{5}5^{4}\]

\[3 - 2x < 625\]

\[2x > - 622\]

\[x > - 311.\]

\[Имеет\ смысл\ при:\]

\[3 - 2x > 0\]

\[2x < 3\]

\[x < 1,5.\]

\[Ответ:\ \ - 311 < x < - 11;\ \ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }1 < x < 1,5.\]