ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 910

\[\boxed{\mathbf{910}.}\]

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

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

\[+ \log_{|2x + 2|}\left( \frac{5}{9} + 3^{x - 1} \right)\]

\[Область\ определения\ функции:\]

\[\left\{ \begin{matrix} 1 - 9^{x} > 0\ \ \ \ \\ 1 + 3^{x} > 0\ \ \ \ \ \\ \frac{5}{9} + 3^{x - 1} > 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} x < 0 \\ x \in R \\ x \in R \\ \end{matrix} \right.\ \rightarrow x < 0.\]

\[0 < |2x + 2 < 1\]

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

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

\[D(y) = \left( - \frac{3}{2}; - 1 \right) \cup \left( - 1;\ - \frac{1}{2} \right).\]

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

\[< \log_{|2x + 2|}\ \left( \left( 1 + 3^{x} \right)\left( \frac{5}{9} + 3^{x - 1} \right) \right)\]

\[\left( 1 - 9^{x} \right) > \left( 1 + 3^{x} \right)\left( \frac{5}{9} + 3^{x - 1} \right)\]

\[\left( 1 - 9^{x} \right) > \frac{5}{9} + 3^{x - 1} +\]

\[+ \frac{5}{9} \cdot 3^{x} + 3^{x} \cdot 3^{x - 1}\]

\[\left( 1 - 9^{x} \right) > \frac{5}{9} + \frac{3^{x}}{3} + \frac{5}{9} \cdot 3^{x} + \frac{9^{x}}{3}\]

\[\left( 1 - 9^{x} \right) > \frac{5}{9} + \frac{8}{3} \cdot 3^{x} + \frac{9^{x}}{3}\]

\[1 - 9^{x} - \frac{5}{9} - \frac{8}{9} \cdot 3^{x} - \frac{9^{x}}{3} > 0\]

\[- \frac{4}{3} \cdot 3^{2x} + \frac{4}{9} - \frac{8}{9} \cdot 3^{x} > 0\]

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

\[- \frac{4}{3}t^{2} - \frac{8}{9}t + \frac{4}{9} > 0\ \ \ \ \ \ \ | \cdot ( - 9)\]

\[12t^{2} + 8t - 4 < 0\ \ \ \ \ |\ :4\]

\[3t^{2} + 2t - 1 < 0\]

\[D_{1} = 1 + 3 = 4\]

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

\[\ t_{2} = \frac{- 1 - 3}{3} = - 1.\]

\[(t + 1)\left( t - \frac{1}{3} \right) < 0\]

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

\[\textbf{а)}\ 3^{x} = 0 - не\ подходит.\]

\[\textbf{б)}\ 3^{x} < \frac{1}{3}\]

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

\[x < - 1.\]

\[Ответ:x \in \left( - \frac{3}{2}; - 1 \right).\]