Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 774

\[\boxed{\mathbf{774}\mathbf{.}}\]

\[1)\ \left( 3^{x} + 2^{x} \right)\left( 3^{x} + 3 \bullet 2^{x} \right) =\]

\[= 8 \bullet 6^{x}\]

\[3^{2x} + (3 \bullet 2)^{x} \bullet 3 + (3 \bullet 2)^{x} +\]

\[+ 3 \bullet 2^{2x} = 8 \bullet (3 \bullet 2)^{x}\]

\[3^{2x} - 4 \bullet (3 \bullet 2)^{x} + 3 \bullet 2^{2x} =\]

\[= 0\ \ \ \ \ |\ :2^{2x}\]

\[\left( \frac{3}{2} \right)^{2x} - 4 \bullet \left( \frac{3}{2} \right)^{x} + 3 = 0\]

\[Пусть\ y = \left( \frac{3}{2} \right)^{x}:\]

\[y^{2} - 4y + 3 = 0\]

\[D = 4^{2} - 4 \bullet 3 = 16 - 12 = 4\]

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

\[\ y_{2} = \frac{4 + 2}{2} = 3.\]

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

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

\[x = 0.\]

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

\[\log_{\frac{3}{2}}\left( \frac{3}{2} \right)^{x} = \log_{\frac{3}{2}}3\]

\[x = \log_{1,5}3\]

\[Ответ:\ \ x_{1} = 0;\ \ x_{2} = \log_{1,5}3.\]

\[2)\ \left( 3 \bullet 5^{x} + 2,5 \bullet 3^{x} \right) \bullet\]

\[\bullet \left( 2 \bullet 3^{x} - 2 \bullet 5^{x} \right) = 8 \bullet 15^{x}\]

\[6 \bullet (5 \bullet 3)^{x} - 6 \bullet 5^{2x} + 5 \bullet 3^{2x} -\]

\[- 5 \bullet (5 \bullet 3)^{x} = 8 \bullet (5 \bullet 3)^{x}\]

\[5 \bullet 3^{2x} - 7 \bullet (5 \bullet 3)^{x} - 6 \bullet 5^{2x} =\]

\[= 0\ \ \ \ \ |\ :5^{2x}\]

\[5 \bullet \left( \frac{3}{5} \right)^{2x} - 7 \bullet \left( \frac{3}{5} \right)^{x} - 6 = 0\]

\[Пусть\ y = \left( \frac{3}{5} \right)^{x}:\]

\[5y^{2} - 7y - 6 = 0\]

\[D = 7^{2} + 4 \bullet 5 \bullet 6 =\]

\[= 49 + 120 = 169\]

\[y_{1} = \frac{7 - 13}{2 \bullet 5} = - \frac{6}{10};\]

\[y_{2} = \frac{7 + 13}{2 \bullet 5} = \frac{20}{10} = 2.\]

\[1)\ \left( \frac{3}{5} \right)^{x} = - \frac{6}{10}\]

\[нет\ корней.\]

\[2)\ \left( \frac{3}{5} \right)^{x} = 2\]

\[\log_{\frac{3}{5}}\left( \frac{3}{5} \right)^{x} = \log_{\frac{3}{5}}2\]

\[x = \log_{0,6}2.\]

\[Ответ:\ \ x = \log_{0,6}2.\]