ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1406

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Год:2020-2021-2022-2023
Тип:учебник
Серия:Базовый и углубленный уровни

Задание 1406

\[\boxed{\mathbf{1406}\mathbf{.}}\]

\[1)\ {3,3}^{x^{2} + 6x} < 1\]

\[{3,3}^{x^{2} + 6x} < {3,3}^{0}\]

\[x^{2} + 6x < 0\]

\[(x + 6) \bullet x < 0\]

\[- 6 < x < 0.\]

\[Ответ:\ \ - 6 < x < 0.\]

\[2)\ \left( \frac{1}{4} \right)^{x - x^{2}} > \frac{1}{2}\]

\[\left( \frac{1}{2} \right)^{2x - 2x^{2}} > \frac{1}{2}\]

\[2x - 2x^{2} < 1\]

\[2x^{2} - 2x + 1 > 0\]

\[D = 4 - 8 = - 4 < 0\]

\[a = 2 > 0;\]

\[x - любое\ число.\]

\[Ответ:\ \ x \in R.\]

\[3)\ {8,4}^{\frac{x - 3}{x^{2} + 6x + 11}} < 1\]

\[{8,4}^{\frac{x - 3}{x^{2} + 6x + 11}} < {8,4}^{0}\]

\[\frac{x - 3}{x^{2} + 6x + 11} < 0\]

\[x^{2} + 6x + 11 = 0\]

\[D = 36 - 44 = - 8 < 0\]

\[a = 1 > 0;\]

\[x^{2} + 6x + 11 > 0.\]

\[Получим:\]

\[x - 3 < 0\]

\[x < 3.\]

\[Ответ:\ \ x < 3.\]

\[4)\ 2^{2x + 1} - 21 \bullet \left( \frac{1}{2} \right)^{2x + 3} + 2 \geq 0\]

\[2^{2x + 1} - \frac{21}{2^{3}} \bullet \left( \frac{1}{2} \right)^{2x} + 2 \geq 0\]

\[2^{2x + 1} - \frac{21}{8 \bullet 2^{2x}} + 2 \geq 0\]

\[y = 2^{2x}:\]

\[2y - \frac{21}{8y} + 2 \geq 0\ \ \ \ \ | \bullet 8y\]

\[16y^{2} + 16y - 21 \geq 0\]

\[D = 256 + 1344 = 1600\]

\[y_{1} = \frac{- 16 - 40}{2 \bullet 16} = - \frac{56}{32} = - \frac{7}{4};\]

\[y_{2} = \frac{- 16 + 40}{2 \bullet 16} = \frac{24}{32} = \frac{3}{4};\]

\[16\left( y + \frac{7}{4} \right)\left( y - \frac{3}{4} \right) \geq 0\]

\[y \leq - \frac{7}{4}\text{\ \ }и\ \ y \geq \frac{3}{4}.\]

\[1)\ 2^{2x} \leq - \frac{7}{4}\]

\[корней\ нет.\]

\[2)\ 2^{2x} \geq \frac{3}{4}\]

\[2x \geq \log_{2}\frac{3}{4}\]

\[x \geq \frac{1}{2}\log_{2}\frac{3}{4}.\]

\[Ответ:\ \ x \geq \frac{1}{2}\log_{2}\frac{3}{4}.\]

\[5)\ 3^{4 - 3x} - 35 \bullet \left( \frac{1}{3} \right)^{2 - 3x} + 6 \geq 0\]

\[\frac{3^{4}}{3^{3x}} - 35 \bullet 3^{3x - 2} + 6 \geq 0\]

\[\frac{81}{3^{3x}} - 35 \bullet \frac{3^{3x}}{3^{2}} + 6 \geq 0\]

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

\[\frac{81}{y} - \frac{35}{9} \bullet y + 6 \geq 0\ \ \ \ \ | \bullet 9y\]

\[729 - 35y^{2} + 54y \geq 0\]

\[35y^{2} - 54y - 729 \leq 0\]

\[D = 2916 + 102\ 060 = 104\ 976\]

\[y_{1} = \frac{54 - 324}{2 \bullet 35} = - \frac{270}{70} = - \frac{27}{7};\]

\[y_{2} = \frac{54 + 324}{2 \bullet 35} = \frac{378}{70} = \frac{27}{5};\]

\[35\left( y + \frac{27}{7} \right)\left( y - \frac{27}{5} \right) \leq 0\]

\[- \frac{27}{7} \leq y \leq 5,4.\]

\[1)\ 3^{3x} \geq - \frac{27}{7}\]

\[корней\ нет.\]

\[2)\ 3^{3x} \leq \frac{27}{5}\]

\[3x \leq \log_{3}\frac{27}{5}\]

\[3x \leq \log_{3}27 - \log_{3}5\]

\[3x \leq 3 - \log_{3}5\]

\[x \leq 1 - \frac{1}{3}\log_{3}5.\]

\[Ответ:\ \ x \leq 1 - \frac{1}{3}\log_{3}5.\]

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