Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 720

\[\boxed{\mathbf{720}.}\]

\[1)\ \left\{ \begin{matrix} 5^{x} - 5^{y} = 100\ \ \ \ \ \ \\ 5^{x - 1} + 5^{y - 1} = 30 \\ \end{matrix} \right.\ \]

\[Пусть\ u = 5^{x};\ \ \text{\ \ }z = 5^{y}:\]

\[\left\{ \begin{matrix} u - z = 100\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ u \bullet 5^{- 1} + z \bullet 5^{- 1} = 30 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} u = 100 + z \\ \frac{u}{5} + \frac{z}{5} = 30\ \ \\ \end{matrix} \right.\ \]

\[1)\ \frac{100 + z}{5} + \frac{z}{5} = 30\]

\[100 + z + z = 150\]

\[2z = 50\ \]

\[z = 25\ \]

\[u = 100 + 25 = 125.\]

\[2)\ 5^{x} = 125\]

\[5^{x} = 5^{3}\ \]

\[x = 3.\]

\[3)\ 5^{y} = 25\]

\[5^{y} = 5^{2}\ \]

\[y = 2.\]

\[Ответ:\ \ (3;\ \ 2).\]

\[2)\ \left\{ \begin{matrix} 2^{x} - 9 \bullet 3^{y} = 7 \\ 2^{x} \bullet 3^{y} = \frac{8}{9}\text{\ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[Пусть\ u = 2^{x};\ \ \ z = 3^{y}:\]

\[\left\{ \begin{matrix} u - 9z = 7 \\ uz = \frac{8}{9}\text{\ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} u = 7 + 9z \\ 9uz = 8\ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[1)\ 9z(7 + 9z) = 8\]

\[63z + 81z^{2} = 8\]

\[81z^{2} + 63z - 8 = 0\]

\[D = 63^{2} + 4 \bullet 81 \bullet 8 =\]

\[= 3969 + 2592 = 6561\]

\[z_{1} = \frac{- 63 - 81}{2 \bullet 81} = - \frac{144}{162} = - \frac{8}{9};\]

\[z_{2} = \frac{- 63 + 81}{2 \bullet 81} = \frac{18}{162} = \frac{1}{9}.\]

\[u_{1} = 7 + 9 \bullet \left( - \frac{8}{9} \right) =\]

\[= 7 - 8 = - 1;\ \]

\[u_{2} = 7 + 9 \bullet \frac{1}{9} = 7 + 1 = 8.\]

\[Уравнения\ имеют\ корни\ при:\]

\[\ z = \frac{1}{9} > 0\ \ и\ \ u = 8 > 0.\]

\[2)\ 2^{x} = 8\]

\[2^{x} = 2^{3}\ \]

\[x = 3.\]

\[3)\ 3^{y} = \frac{1}{9}\]

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

\[y = - 2\ \]

\[Ответ:\ \ (3; - 2).\]

\[3)\ \left\{ \begin{matrix} 16^{y} - 16^{x} = 24 \\ 16^{x + y} = 256\ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Пусть\ u = 16^{x};\ \text{\ \ }z = 16^{y}:\]

\[\left\{ \begin{matrix} z - u = 24 \\ zu = 256\ \ \ \\ \end{matrix} \right.\ \text{\ \ }\]

\[\left\{ \begin{matrix} z = 24 + u\ \ \ \ \ \\ zu - 256 = 0 \\ \end{matrix} \right.\ \]

\[(24 + u)u - 256 = 0\]

\[u^{2} + 24u - 256 = 0\]

\[D = 24^{2} + 4 \bullet 256 =\]

\[= 576 + 1024 = 1600\]

\[u_{1} = \frac{- 24 - 40}{2} = - 32;\ \text{\ \ }\]

\[u_{2} = \frac{- 24 + 40}{2} = 8.\]

\[z_{1} = 24 - 32 = - 8;\ \ \ \ \ \ \ \ \ \]

\[\ z_{2} = 24 + 8 = 32.\]

\[Уравнения\ имеют\ корни\ при:\]

\[u = 8 > 0\ \ и\ \ z = 32 > 0.\]

\[1)\ 16^{x} = 8\]

\[2^{4x} = 2^{3}\]

\[4x = 3\]

\[x = 0,75.\]

\[2)\ 16^{y} = 32\]

\[2^{4y} = 2^{5}\]

\[4y = 5\]

\[y = 1,25.\]

\[Ответ:\ \ (0,75;1,25).\]

\[4)\ \left\{ \begin{matrix} 3^{x} + 2^{x + y + 1} = 5 \\ 3^{x + 1} - 2^{x + y} = 1 \\ \end{matrix} \right.\ \]

\[Пусть\ u = 3^{x};\ \text{\ \ }z = 2^{x + y}:\]

\[\left\{ \begin{matrix} u + z \bullet 2 = 5 \\ u \bullet 3 - z = 1 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\left\{ \begin{matrix} u = 5 - 2z \\ 3u - z = 1 \\ \end{matrix} \right.\ \]

\[3(5 - 2z) - z = 1\]

\[15 - 6z - z = 1\]

\[- 7z = - 14\ \]

\[z = 2.\]

\[u = 5 - 2 \bullet 2 = 5 - 4 = 1.\]

\[1)\ 3^{x} = 1\]

\[3^{x} = 3^{0}\ \]

\[x = 0.\]

\[2)\ 2^{x + y} = 2\]

\[2^{y} = 2^{1}\ \]

\[y = 1.\]

\[Ответ:\ \ (0;1).\]