Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 262

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\[1)\ \left\{ \begin{matrix} 2^{x - y} = 128\ \ \ \ \ \\ \left( \frac{1}{2} \right)^{x - 2y + 1} = \frac{1}{8} \\ \end{matrix} \right.\ \ \]

\[\left\{ \begin{matrix} 2^{x - y} = 2^{7}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \left( \frac{1}{2} \right)^{x - 2y + 1} = \left( \frac{1}{2} \right)^{3} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x - y = 7\ \ \ \ \ \ \ \ \ \ \\ x - 2y + 1 = 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} x = 7 + y\ \ \\ x = 2y + 2 \\ \end{matrix} \right.\ \ \]

\[7 + y = 2y + 2\ \]

\[- y = - 5\ \]

\[y = 5;\ \]

\[x = 7 + 5 = 12.\ \]

\[Ответ:\ \ (12;\ \ 5).\]

\[2)\ \left\{ \begin{matrix} 2^{x} \bullet 5^{y} = 10 \\ 5^{y} - 2^{x} = 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} 2^{x} \bullet 5^{y} - 10 = 0 \\ 5^{y} = 2^{x} + 3\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \ \]

\[2^{x} \bullet \left( 2^{x} + 3 \right) - 10 = 0\ \]

\[2^{2x} + 3 \bullet 2^{x} - 10 = 0\ \]

\[Пусть\ z = 2^{x}:\]

\[z^{2} + 3z - 10 = 0\ \]

\[D = 3^{2} + 4 \bullet 10 = 9 + 40 = 49\]

\[z_{1} = \frac{- 3 - 7}{2} =\]

\[= - 5\ (не\ подходит);\text{\ \ }\]

\[z_{2} = \frac{- 3 + 7}{2} = 2.\ \]

\[1)\ 2^{x} = 2\]

\[x = 1.\ \]

\[2)\ 5^{y} = 2 + 3\ \]

\[5^{y} = 5\ \]

\[y = 1.\ \]

\[Ответ:\ \ (1;\ \ 1).\]