Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 245

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\[1)\ \left\{ \begin{matrix} \left( 5^{x} \right)^{y} = 5^{21}\text{\ \ } \\ 5^{x} \bullet 5^{y} = 5^{10} \\ 3^{x} > 3^{y}\text{\ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 5^{\text{xy}} = 5^{21}\text{\ \ } \\ 5^{x + y} = 5^{10} \\ x > y\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} xy = 21\ \ \ \ \ \\ x + y = 10 \\ x > y\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} xy - 21 = 0 \\ y = 10 - x\ \ \\ x > y\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[1)\ x(10 - x) - 21 = 0\]

\[10x - x^{2} - 21 = 0\]

\[x^{2} - 10x + 21 = 0\]

\[D = 10^{2} - 4 \bullet 21 =\]

\[= 100 - 84 = 16\]

\[x_{1} = \frac{10 - 4}{2} = 3;\ \ \ \text{\ \ }\]

\[x_{2} = \frac{10 + 4}{2} = 7\ \]

\[y_{1} = 10 - 3 = 7;\ \ \ \ \ \text{\ \ }\]

\[y_{2} = 10 - 7 = 3.\]

\[Ответ:\ \ (7;3).\]

\[2)\ \left\{ \begin{matrix} \left( {0,2}^{y} \right)^{x} = 0,008 \\ {0,4}^{y} = {0,4}^{3,5 - x}\text{\ \ } \\ 2^{x} \bullet {0,5}^{y} > 1\ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} {0,2}^{\text{xy}} = {0,2}^{3}\text{\ \ \ \ \ \ \ \ \ \ \ \ } \\ {0,4}^{y} = {0,4}^{3,5}\ :{0,4}^{x} \\ 2^{x} \bullet \left( \frac{1}{2} \right)^{y} > 1\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} xy = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ {0,4}^{y} \bullet {0,4}^{x} = {0,4}^{3,5} \\ 2^{x} \bullet 2^{- y} > 1\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} xy = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ {0,5}^{y + x} = {0,4}^{3,5} \\ 2^{x - y} > 2^{0}\text{\ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \ \]

\[\left\{ \begin{matrix} xy = 3\ \ \ \ \ \ \ \ \\ x + y = 3,5 \\ x - y > 0\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} xy - 3 = 0\ \\ y = 3,5 - x \\ x > y\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[1)\ x(3,5 - x) - 3 = 0\]

\[3,5x - x^{2} - 3 = 0\ \ \ \ \ \ \ \ \ \ | \bullet ( - 2)\]

\[2x^{2} - 7x + 6 = 0\]

\[D = 7^{2} - 4 \bullet 2 \bullet 6 = 49 - 48 =\]

\[x_{1} = \frac{7 - 1}{2 \bullet 2} = \frac{6}{4} = 1,5;\ \ \ \ \text{\ \ }\]

\[x_{2} = \frac{7 + 1}{2 \bullet 2} = \frac{8}{4} = 2\ \]

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

\[y_{2} = 3,5 - 2 = 1,5.\]

\(Ответ:\ \ (2;1,5).\)