ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1575

\[\boxed{\mathbf{1575}\mathbf{.}}\]

\[2x + 2 > 0\]

\[x + 1 > 0\]

\[x > - 1.\]

\[x > - 1:\]

\[2x + 2 > 1\]

\[2x > - 1\]

\[x > - \frac{1}{2}.\]

\[x < - 1:\ \]

\[- (2x + 2) > 1\]

\[- 2x - 2 > 1\]

\[2x < - 3\]

\[x < - \frac{3}{2}.\]

\[x < - \frac{3}{2};\ x > - \frac{1}{2}:\]

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

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

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

\[9 - 9 \bullet 3^{x} - 9 \bullet 3^{x - 1} < 5\]

\[- 3^{2 + x} - 3^{2 + x - 1} < - 4\]

\[3^{x + 2} + 3^{x + 1} > 4\]

\[3^{x} \bullet (9 + 3) > 4\]

\[3^{x} \bullet 12 > 4\]

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

\[x > - 1.\]

\[- \frac{3}{2} < x < - \frac{1}{2}:\]

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

\[x < - 1.\]

\[Имеет\ смысл\ при:\]

\[1 - 9^{x} > 0\]

\[9^{x} < 1\]

\[9^{x} < 9^{0}\]

\[x < 0.\]

\[Ответ:\ \ - \frac{3}{2} < x < - 1;\ \ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } - \frac{1}{2} < x < 0.\]