Решебник по алгебре 9 класс Мерзляк Задание 861

 

\[\boxed{\mathbf{861\ (861).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[Пусть\ числа\ a_{1},\ a_{2},\ a_{3}.\]

\[Получаем:\]

\[a_{1} + a_{2} + a_{3} = 21\ \]

\[a_{1} + a_{1} + d + a_{1} + 2d = 21\ \ \]

\[3a_{1} + 3d = 21\ \ \ |\ :3,\ \ \]

\[a_{1} + d = 7\ \]

\[\ d = 7 - a_{1}.\]

\[Значит,\ образуют\ \]

\[геометрическую\ \]

\[прогрессию\ числа\ \]

\[a_{1} + 2,a_{2} + 3,a_{3} + 9.\]

\[Отсюда:\]

\[\ \left( a_{2} + 3 \right)^{2} = \left( a_{1} + 2 \right)\left( a_{3} + 9 \right)\]

\[\left( a_{1} + d + 3 \right)^{2} =\]

\[= \left( a_{1} + 2 \right)\left( a_{1} + 2d + 9 \right)\]

\[(7 + 3)^{2} =\]

\[= \left( a_{1} + 2 \right)\left( a_{1} + 2 \cdot \left( 7 - a_{1} \right) + 9 \right),\]

\[10^{2} =\]

\[= \left( a_{1} + 2 \right)\left( a_{1} + 14 - 2a_{1} + 9 \right)\ \]

\[10^{2} = \left( a_{1} + 2 \right)\left( 23 - a_{1} \right),\]

\[100 = 23a_{1} - a_{1}^{2} + 46 - 2a_{1}\text{\ \ }\]

\[a_{1}^{2} - 21a_{1} + 54 = 0\]

\[a_{1} + a_{2} = 21;\ \ a_{1} = 18\]

\[a_{1}a_{2} = 54;\ \ a_{1} = 3\]

\[при\ \ a_{1} = 3 \Longrightarrow \ \ d = 7 - 3 = 4:\]

\[a_{2} = 3 + 4 = 7;\ \]

\[a_{3} = a_{2} + d = 7 + 4 = 11;\]

\[при\ a_{1} = 18 \Longrightarrow d =\]

\[= 7 - 18 = - 11:\]

\[a_{2} = 18 - 11 = 7;\ \]

\[a_{3} = 7 - 11 =\]

\[= - 4 < 0 \Longrightarrow не\]

\[\ удовлетворяет.\]

\[Ответ:3;7;11.\]

\[\boxed{\mathbf{861.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[b_{1} + b_{4} = \frac{35}{3},\ \ b_{2} + b_{3} = 10\]

\[\left\{ \begin{matrix} b_{1} + b_{1}q^{3} = \frac{35}{3} \\ b_{1}q + b_{1}q^{2} = 10 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} b_{1}\left( 1 + q^{3} \right) = \frac{35}{3}\text{\ \ \ \ \ } \\ b_{1}q(1 + q) = 10\ \ \ \\ \end{matrix} \right.\ \ (\ :)\ \ \]

\[\frac{b_{1}(1 + q)\left( 1 - q + q^{2} \right)}{b_{1}q(1 + q)} = \frac{35}{30}\]

\[\frac{1 - q + q^{2}}{q} = \frac{7}{6}\]

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

\[6q^{2} - 13q + 6 = 0\]

\[D = 169 - 144 = 25\]

\[q = \frac{13 + 5}{12} = 1,5;\ \ \ \ \ \ \ \]

\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }q = \frac{13 - 5}{12} = \frac{2}{3}\]

\[при\ \ q = 1,5:\ \ \]

\[b_{1} = \frac{10}{1,5 \cdot (1 + 1,5)} = \frac{10}{1,5 \cdot 2,5} =\]

\[= \frac{10}{\frac{3}{2} \cdot \frac{5}{2}} = \frac{8}{3} = 2\frac{2}{3};\]

\[b_{2} = \frac{8}{3} \cdot \frac{3}{2} = 4,\ \ \]

\[b_{3} = 4 \cdot \frac{3}{2} = 6,\]

\[\text{\ \ }b_{4} = 6 \cdot \frac{3}{2} = 9.\]

\[при\ q = \frac{2}{3}:1 = 1023 \cdot 1 + \frac{2}{3} =\]

\[= 102 \cdot 53 \cdot 3 = 9,\]

\[\ \ b2 = 9 \cdot 23 = 6,\]

\[b_{3} = 6 \cdot \frac{2}{3} = 4,\ \ \]

\[b_{4} = 4 \cdot \frac{2}{3} = \frac{8}{3}.\]

\[Ответ:\ \ 2\frac{2}{3};4;6;9.\]