Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 366

\[\boxed{\mathbf{366}\mathbf{.}}\]

\[\frac{2}{3^{x} - 1} \leq \frac{7}{9^{x} - 2}\]

\[\frac{2}{3^{x} - 1} - \frac{7}{9^{x} - 2} \leq 0\]

\[\frac{2\left( 9^{x} - 2 \right) - 7\left( 3^{x} - 1 \right)}{\left( 3^{x} - 1 \right)\left( 9^{x} - 2 \right)} \leq 0\]

\[\frac{2 \bullet 9^{x} - 4 - 7 \bullet 3^{x} + 7}{\left( 3^{x} - 1 \right)\left( 9^{x} - 2 \right)} \leq 0\]

\[\frac{2 \bullet 3^{2x} - 7 \bullet 3^{x} + 3}{\left( 3^{x} - 1 \right)\left( 3^{2x} - 2 \right)} \leq 0\]

\[2 \bullet 3^{2x} - 7 \bullet 3^{x} + 3 = 0\]

\[D = 7^{2} - 4 \bullet 2 \bullet 3 = 49 - 24 =\]

\[= 25,\ тогда:\]

\[3_{1}^{x} = \frac{7 - 5}{2 \bullet 2} = \frac{2}{4} = \frac{1}{2}\]

\[3_{2}^{x} = \frac{7 + 5}{2 \bullet 2} = \frac{12}{4} = 3\]

\[\left( 3^{x} - \frac{1}{2} \right)\left( 3^{x} - 3 \right) = 0\]

\[\frac{\left( 3^{x} - \frac{1}{2} \right)\left( 3^{x} - 3 \right)}{\left( 3^{x} - 1 \right)\left( 3^{2x} - 2 \right)} \leq 0\]

\[x < - \sqrt{2};\ \ \ \]

\[\ \frac{1}{2} \leq 3^{x} < 1;\ \ \ \]

\[\ \sqrt{2} < 3^{x} \leq 3.\]

\[1)\ 3^{x} < - \sqrt{2}\]

\[нет\ корней.\]

\[2)\ 3^{x} \geq \frac{1}{2}\]

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

\[x \geq \log_{3}2^{- 1}\]

\[\ x \geq - \log_{3}2.\]

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

\[3^{x} < 3^{0}\ \]

\[x < 0.\]

\[4)\ 3^{x} > \sqrt{2}\]

\[\log_{3}3^{x} > \log_{3}\sqrt{2}\ \]

\[x > \log_{3}\sqrt{2}.\]

\[5)\ 3^{x} \leq 3\]

\[x \leq 1.\]

\[Ответ:\ \ - \log_{3}2 \leq x < 0;\ \ \ \]

\[\log_{3}\sqrt{2} < x \leq 1.\]