ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 909

\[1)\ 3^{\log_{2}\frac{x - 1}{x + 2}} < \frac{1}{9}\]

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

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

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

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

\[\frac{4(x - 1) - (x + 2)}{4(x + 2)} < 0\]

\[\frac{4x - 4 - x - 2}{x + 2} < 0\]

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

\[- 2 < x < 2.\]

\[Область\ определения:\]

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

\[x < - 2;\text{\ \ }\ x > 1.\]

\[Ответ:\ \ x \in (1;\ 2).\]

\[2)\ 5^{\log_{2}\left( x^{2} - 4x + 3,5 \right)} > \frac{1}{5}\]

\[5^{\log_{2}\left( x^{2} - 4x + 3,5 \right)} > 5^{- 1}\]

\[\log_{2}\left( x^{2} - 4x + 3,5 \right) > - 1\]

\[x^{2} - 4x + 3,5 > 2^{- 1}\]

\[x^{2} - 4x + 3,5 > 0,5\]

\[x^{2} - 4x + 3 > 0\]

\[D = 16 - 12 = 4\]

\[x_{1} = \frac{4 - 2}{2} = 1;\]

\[x_{2} = \frac{4 + 2}{2} = 3;\]

\[(x - 1)(x - 3) > 0\]

\[x < 1;\ \text{\ \ }x > 3.\]

\[Ответ:\ \ x \in ( - \infty;\ 1) \cup (3;\ + \infty).\]

\[3)\ 2^{\log_{0,7}(1 + 2x)} > 4\]

\[2^{\log_{0,7}(1 + 2x)} > 2^{2}\]

\[\log_{0,7}(1 + 2x) > 2\]

\[0 < 1 + 2x < {0,7}^{2}\]

\[0 < 1 + 2x < 0,49\]

\[- 1 < 2x < - 0,51\]

\[- 0,5 < x < - 0,255.\]

\(Ответ:\ \ x \in ( - 0,5;\ - 0,255).\)