Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 240

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\[1)\ \left\{ \begin{matrix} 2x - y = 1 \\ 5^{x + y} = 25\ \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} y = 2x - 1 \\ 5^{x + y} = 5^{2}\text{\ \ } \\ \end{matrix} \right.\ \]

\[5^{x + 2x - 1} = 5^{2}\]

\[x + 2x - 1 = 2\]

\[3x = 3\]

\[\ x = 1.\]

\[y = 2 \bullet 1 - 1 = 2 - 1 = 1.\]

\[Ответ:\ \ (1;\ \ 1).\]

\[2)\ \left\{ \begin{matrix} x - y = 2 \\ 3^{x^{2} + y} = \frac{1}{9} \\ \end{matrix} \right.\ \]

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

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

\[(2 + y)^{2} + y = - 2\]

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

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

\[D = 5^{2} - 4 \bullet 6 = 25 - 24 = 1\]

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

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

\[x_{1} = 2 - 3 = - 1;\ \ \ \ \ \ \text{\ \ }\]

\[x_{2} = 2 - 2 = 0\ \]

\[Ответ:\ \ ( - 1; - 3);\ \ \ (0; - 2).\]

\[3)\ \left\{ \begin{matrix} x + y = 1 \\ 2^{x - y} = 8\ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]

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

\[2^{x - (1 - x)} = 2^{3}\]

\[x - (1 - x) = 3\]

\[x - 1 + x = 3\]

\[2x = 4\]

\[x = 2.\]

\[y = 1 - 2 = - 1.\]

\[Ответ:\ \ (2;\ - 1).\]

\[4)\ \left\{ \begin{matrix} x + 2y = 3 \\ 3^{x - y} = 81\ \\ \end{matrix} \right.\ \ \]

\[\left\{ \begin{matrix} x = 3 - 2y \\ 3^{x - y} = 3^{4}\text{\ \ } \\ \end{matrix} \right.\ \]

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

\[3 - 2y - y = 4\]

\[- 3y = 1\ \]

\[y = - \frac{1}{3}.\]

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

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