Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 923

\[1)\ \sqrt{32^{x} + 4} - \sqrt{\left| 32^{x} - 7 \right|} < 1\]

\[t = 32^{x}:\]

\[\sqrt{t + 4} - \sqrt{|t - 7|} < 1\]

\[\sqrt{t + 4} < 1 + \sqrt{|t - 7|}\]

\[t + 4 < 1 + 2\sqrt{|t - 7|} + |t - 7|.\]

\[t \geq 7:\]

\[t + 4 < 1 + 2\sqrt{t - 7} + t - 7\]

\[2\sqrt{t - 7} > 10\]

\[\sqrt{t - 7} > 5\]

\[t - 7 > 25\]

\[t > 32\]

\[32^{x} > 32\]

\[x > 1.\]

\[t \leq 7:\]

\[t + 4 < 1 + 2\sqrt{7 - t} + 7 - t\]

\[2\sqrt{7 - t} > 2t - 4\]

\[\sqrt{7 - t} > t - 2\]

\[7 - t > t^{2} - 4t + 4\]

\[t^{2} - 3t - 3 < 0\]

\[D = 3^{2} + 4 \bullet 3 = 9 + 12 = 21\]

\[t = \frac{3 \pm \sqrt{21}}{2};\]

\[\frac{3 - \sqrt{21}}{2} < t < \frac{3 + \sqrt{21}}{2}\]

\[\frac{3 - \sqrt{21}}{2} < 32^{x} < \frac{3 + \sqrt{21}}{2}\]

\[x < \log_{32}\frac{3 + \sqrt{21}}{2}.\]

\[Ответ:\ \ \]

\[x \in \left( - \infty;\ \log_{32}\frac{3 + \sqrt{21}}{2} \right) \cup (1;\ + \infty).\]

\[2)\ 3^{x}\left( \sqrt{9^{1 - x} - 1} + 1 \right) < 3\left| 3^{x} - 1 \right|\]

\[\sqrt{3^{2x}\left( \frac{9}{3^{2x}} - 1 \right)} + 3^{x} < 3\left| 3^{x} - 1 \right|\]

\[\sqrt{9 - 3^{2x}} + 3^{x} < 3\left| 3^{x} - 1 \right|\]

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

\[\sqrt{9 - t^{2}} + t < 3|t - 1|.\]

\[t \geq 1:\]

\[\sqrt{9 - t^{2}} + t < 3(t - 1)\]

\[\sqrt{9 - t^{2}} < 2t - 3\]

\[9 - t^{2} < 4t^{2} - 12t + 9\]

\[5t^{2} - 12t > 0\]

\[t(5t - 12) > 0\]

\[t < 0;\ \ \ t > \frac{12}{5};\]

\[3^{x} > \frac{12}{5}\]

\[x > \log_{3}\frac{12}{5}.\]

\[t \leq 1:\]

\[\sqrt{9 - t^{2}} + t < 3(1 - t)\]

\[\sqrt{9 - t^{2}} < 3 - 4t\]

\[9 - t^{2} < 9 - 24t + 16t^{2}\]

\[17t^{2} - 24t > 0\]

\[t(17t - 24) > 0\]

\[t < 0;\ \ \ t > \frac{24}{17}.\]

\[Область\ определения:\]

\[9 - t^{2} \geq 0\]

\[(t + 3)(t - 3) \leq 0\]

\[- 3 \leq t \leq 3\]

\[- 3 \leq 3^{x} \leq 3\]

\[x \leq 1.\]

\[Ответ:\ \ x \in \left( \log_{3}\frac{12}{5};\ 1 \right\rbrack.\]