ГДЗ по алгебре 9 класс Макарычев Задание 644

\[\boxed{\text{644\ (644).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[x_{1} = a,\ \ x_{2} = a + d,\ \ x_{3} = a + 2d,\ \ x_{1} + 1 = a + 1,\]

\[x_{2} + 1 = (a + 1)q,\ \ x_{3} + 4 = (a + 1)q^{2},\]

\[\left\{ \begin{matrix} a + d = (a + 1)q - 1\ \ \ \ \ \\ a + 2d = (a + 1)q^{2} - 4 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} d = (a + 1)(a - 1)\text{\ \ \ \ \ \ \ \ \ \ \ \ } \\ 2d + 3 = (a + 1)\left( q^{2} - 1 \right) \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 2d + 3 = \frac{d}{q - 1} \cdot \left( q^{2} - 1 \right) \\ q - 1 = \frac{d}{a + 1}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} 2 + \frac{3}{d} = q + 1 \\ a + 1 = \frac{d}{q - 1} \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} q = \frac{3}{d} + 1\ \ \ \ \ \ \ \\ a = \frac{d}{q - 1} - 1 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} q = \frac{3 + d\ }{d}\text{\ \ } \\ a = \frac{d^{2}}{3} - 1 \\ \end{matrix} \right.\ \]

\[x_{1} + x_{2} + x_{3} = 3a + 3d = 15,\ \ a + d = 5,\]

\[\frac{d^{2}}{3} - 1 + d = 15,\ \ d^{2} + 3d - 3 = 15,\]

\[d^{2} + 3d - 18 = 0,\ \ по\ теореме\ Виета:\ \ d_{1} = - 6,\ \ d_{2} = 3,\]

\[1)\ d_{1} = - 6,\ \ a + d = 5,\ \ x_{1} = a = 11,\]

\[x_{2} = 5,\ \ x_{3} = - 1 \Longrightarrow не\ подходит,\ так\ как\ по\ условию\ все\ числа\]

\[положительные;\]

\[2)\ d_{2} = 3,\ \ a + d = 5,\ \ x_{1} = a = 2,\ \ x_{2} = 5,\ \ x_{3} = 8.\]

\[Ответ:2;5;8.\]