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МатематикаРешебник по алгебре 9 класс Макарычев Задание 644
Задание 644
\[\boxed{\text{644\ (644).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[x_{1} = a,\ \ x_{2} = a + d,\ \ x_{3} = a + 2d,\ \ x_{1} + 1 = a + 1,\]
\[x_{2} + 1 = (a + 1)q,\ \ x_{3} + 4 = (a + 1)q^{2},\]
\[\left\{ \begin{matrix} a + d = (a + 1)q - 1\ \ \ \ \ \\ a + 2d = (a + 1)q^{2} - 4 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} d = (a + 1)(a - 1)\text{\ \ \ \ \ \ \ \ \ \ \ \ } \\ 2d + 3 = (a + 1)\left( q^{2} - 1 \right) \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} 2d + 3 = \frac{d}{q - 1} \cdot \left( q^{2} - 1 \right) \\ q - 1 = \frac{d}{a + 1}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} 2 + \frac{3}{d} = q + 1 \\ a + 1 = \frac{d}{q - 1} \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} q = \frac{3}{d} + 1\ \ \ \ \ \ \ \\ a = \frac{d}{q - 1} - 1 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} q = \frac{3 + d\ }{d}\text{\ \ } \\ a = \frac{d^{2}}{3} - 1 \\ \end{matrix} \right.\ \]
\[x_{1} + x_{2} + x_{3} = 3a + 3d = 15,\ \ a + d = 5,\]
\[\frac{d^{2}}{3} - 1 + d = 15,\ \ d^{2} + 3d - 3 = 15,\]
\[d^{2} + 3d - 18 = 0,\ \ по\ теореме\ Виета:\ \ d_{1} = - 6,\ \ d_{2} = 3,\]
\[1)\ d_{1} = - 6,\ \ a + d = 5,\ \ x_{1} = a = 11,\]
\[x_{2} = 5,\ \ x_{3} = - 1 \Longrightarrow не\ подходит,\ так\ как\ по\ условию\ все\ числа\]
\[положительные;\]
\[2)\ d_{2} = 3,\ \ a + d = 5,\ \ x_{1} = a = 2,\ \ x_{2} = 5,\ \ x_{3} = 8.\]
\[Ответ:2;5;8.\]
\[\boxed{\text{644.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ a_{1} = 9\sqrt{3} - 2,\ \ d = 2 - \sqrt{3},\]
\[a_{12} = a_{1} + d(12 - 1) =\]
\[= a_{1} + 11d =\]
\[= 9\sqrt{3} - 2 + 22 - 11\sqrt{3} =\]
\[= 20 - 2\sqrt{3}.\]
\[\textbf{б)}\ a_{1} = \frac{5\sqrt{3} - 7}{3},\]
\[\ \ d = \frac{\sqrt{3} - 2}{3},\]
\[a_{8} = a_{1} + d(8 - 1) = a_{1} + 7d =\]
\[= \frac{5\sqrt{3} - 7}{3} + 7 \cdot \frac{\sqrt{3} - 2}{3} =\]
\[= \frac{12\sqrt{3} - 21}{3} = 4\sqrt{3} - 7.\]
