ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 1607

\[\boxed{\mathbf{1607}\mathbf{.}}\]

\[1)\ \left\{ \begin{matrix} 6^{x} - 2 \bullet 3^{y} = 2 \\ 6^{x} \bullet 3^{y} = 12\ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} 6^{x} = 2 \bullet 3^{y} + 2 \\ 6^{x} \bullet 3^{y} = 12\ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left( 2 \bullet 3^{y} + 2 \right) \bullet 3^{y} = 12\]

\[2 \bullet 3^{2y} + 2 \bullet 3^{y} - 12 = 0\]

\[z = 3^{y}:\]

\[2z^{2} + 2z - 12 = 0\]

\[z^{2} + z - 6 = 0\]

\[D = 1 + 24 = 25\]

\[z_{1} = \frac{- 1 - 5}{2} = - 3;\]

\[z_{2} = \frac{- 1 + 5}{2} = 2.\]

\[1)\ 3^{y} = - 3\]

\[корней\ нет.\]

\[2)\ 3^{y} = 2\ \]

\[y = \log_{3}2.\]

\[6^{x} = 2 \bullet 3^{\log_{3}2} + 2\]

\[6^{x} = 2 \bullet 2 + 2\]

\[6^{x} = 4 + 2 = 6\]

\[x = 1.\]

\[Ответ:\ \ \left( 1;\ \log_{3}2 \right).\]

\[2)\ \left\{ \begin{matrix} 7 \bullet 2^{x} + 6y = 2\ \ \ \ \ \ \\ 3 \bullet 2^{x + 1} - 5y = 93 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} 2^{x} = \frac{2 - 6y}{7}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 3 \bullet 2^{x} \bullet 2 - 5y = 93 \\ \end{matrix} \right.\ \]

\[3 \bullet \frac{2 - 6y}{7} \bullet 2 - 5y = 93\ \ \ \ \ | \bullet 7\]

\[6(2 - 6y) - 7 \bullet 5y = 93 \bullet 7\]

\[12 - 36y - 35y = 651\]

\[- 71y = 639\]

\[y = - 9.\]

\[2^{x} = \frac{2 - 6 \bullet ( - 9)}{7}\]

\[2^{x} = \frac{2 + 54}{7}\]

\[2^{x} = \frac{56}{7} = 8\]

\[x = 3.\]

\[Ответ:\ \ (3;\ - 9).\]