Решебник по алгебре 9 класс Мерзляк Задание 864

\[\boxed{\mathbf{864\ (864).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[Геометрическая\ прогрессия:\]

\[b_{1},\ b_{2},\ b_{3};\ \ \ \ \ \ \ \]

\[\ b_{1} + b_{2} + \ b_{3} = 26\]

\[b_{1} + b_{1}q + b_{1}q^{2} = 26\ \ \]

\[b_{1}\left( 1 + q + q^{2} \right) = 26.\]

\[Арифметическая\ прогрессия:\]

\[\text{\ \ }b_{1} + 1,b_{2} + 6,\ b_{3} + 3\]

\[b_{2} + 6 = \frac{b_{1} + 1 + b_{3} + 3}{2}\ \]

\[\ 2 \cdot \left( b_{2} + 6 \right) = b_{1} + b_{3} + 4\]

\[2b_{2} + 12 = b_{1} + b_{3} + 4\]

\[\ b_{1} + b_{3} - 2b_{2} = 8\]

\[b_{1} + b_{1}q^{2} - 2b_{1}q = 8\ \ \]

\[b_{1}\left( 1 + q^{2} - 2q \right) = 8\]

\[Запишем\ систему:\]

\[\left\{ \begin{matrix} b_{1}\left( 1 + q + q^{2} \right) = 26 \\ b_{1}\left( 1 + q^{2} - 2q \right) = 8 \\ \end{matrix}\ \ |(\ :) \right.\ \]

\[\ \frac{b_{1}\left( 1 + q + q^{2} \right)}{b_{1}\left( 1 + q^{2} - 2q \right)} = \frac{26}{8}\]

\[\frac{1 + q + q^{2}}{1 + q^{2} - 2q} = \frac{13}{4}\]

\[4 \cdot \left( 1 + q + q^{2} \right) =\]

\[= 13 \cdot \left( 1 + q^{2} - 2q \right)\]

\[4 + 4q + 4q^{2} =\]

\[= 13 + 13q^{2} - 26q\ \ \]

\[9q^{2} - 30q + 9 = 0\ \ \ |\ :3\]

\[3q^{2} - 10q + 3 = 0\]

\[D = 100 - 36 = 64\]

\[q = \frac{10 + 8}{6} = 3;\ \ \]

\[q = \frac{10 - 8}{6} = \frac{1}{3}\]

\[b_{1}\left( 1 + q^{2} - 2q \right) = 8\ \ \]

\[b_{1}(1 - q)^{2} = 8\]

\[b_{1} = \frac{8}{(1 - q)^{2}}\]

\[при\ \ q = 3:\ \ \]

\[b_{1} = \frac{8}{(1 - 3)^{2}} = \frac{8}{4} = 2,\ \ \]

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

\[b_{3} = 6 \cdot 3 = 18;\]

\[при\ q = \frac{1}{3}:\ \ \]

\[b_{1} = \frac{8}{\left( 1 - \frac{1}{3} \right)^{2}} = \frac{8 \cdot 9}{4} = 18,\]

\[\text{\ \ }b_{2} = 18 \cdot \frac{1}{3} = 6,\]

\[\text{\ \ }b_{3} = 6 \cdot \frac{1}{3} = 2.\]

\[Ответ:2,\ 6,\ 18\ \ \ или\ \ \ 18,\ 6,\ 2.\]