ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 926

\[9^{|x|} - 8 \bullet 3^{x} > 9\]

\[3^{2|x|} - 8 \bullet 3^{x} - 9 > 0.\]

\[1)\ \ x \geq 0:\]

\[3^{2x} - 8 \bullet 3^{x} - 9 > 0\]

\[D = 64 + 36 = 100\]

\[3_{1}^{x} = \frac{8 - 10}{2} = - 1;\]

\[3_{2}^{x} = \frac{8 + 10}{2} = 9;\]

\[\left( 3^{x} + 1 \right)\left( 3^{x} - 9 \right) > 0\]

\[3^{x} > 9\]

\[x > 2.\]

\[2)\ x \leq 0:\]

\[3^{- 2x} - 8 \bullet 3^{x} - 9 > 0\]

\[1 - 8 \bullet 3^{3x} - 9 \bullet 3^{2x} > 0\]

\[8 \bullet 3^{3x} + 9 \bullet 3^{2x} - 1 < 0\]

\[\left( 3^{x} + 1 \right)\left( 8 \bullet 3^{2x} + 3^{x} - 1 \right) < 0\]

\[D = 1^{2} + 4 \bullet 8 = 1 + 32 = 33\]

\[3^{x} = \frac{- 1 \pm \sqrt{33}}{2 \bullet 8} = \frac{- 1 \pm \sqrt{33}}{16}\]

\[3^{x} < - 1;\ \ \]

\[\frac{- 1 - \sqrt{33}}{16} < 3^{x} < \frac{- 1 + \sqrt{33}}{16}\]

\[x < \log_{3}\frac{\sqrt{33} - 1}{16}.\]

\[Ответ:\ \ \]

\[x \in \left( - \infty;\ \log_{3}\frac{\sqrt{33} - 1}{16} \right) \cup (2;\ + \infty);\]

\[x = 3.\]