Решебник по алгебре 9 класс Мерзляк Задание 863

\[\boxed{\mathbf{863\ (863).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[Геометрическая\ прогрессия:\]

\[b_{1},\ b_{2},\ b_{3},\ \ b_{1} + b_{2} + \ b_{3} = 65,\]

\[b_{1} + b_{1}q + b_{1}q^{2} = 65,\ \ \]

\[b_{1}\left( 1 + q + q^{2} \right) = 65.\]

\[Арифметическая\ прогрессия:\ \]

\[b_{1} - 1,b_{2},\ b_{3} - 19\ \]

\[b_{2} = \frac{b_{1} - 1 + b_{3} - 19}{2}\]

\[\ 2b_{2} = b_{1} + b_{3} - 20\]

\[\ 2b_{2} - b_{1} - b_{3} = - 20\]

\[b_{1} + b_{3} - 2b_{2} = 20\]

\[b_{1} + b_{1}q^{2} - 2b_{1}q = 20\ \ \]

\[b_{1}\left( 1 + q^{2} - 2q \right) = 20.\]

\[Запишем\ систему:\]

\[\left\{ \begin{matrix} b_{1}\left( 1 + q^{2} - 2q \right) = 20 \\ b_{1}\left( 1 + q + q^{2} \right) = 65 \\ \end{matrix}\ |\ (\ :)\text{\ \ \ \ } \right.\ \]

\[\frac{b_{1}\left( 1 + q^{2} - 2q \right)}{b_{1}\left( 1 + q + q^{2} \right)} = \frac{20}{65}\]

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

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

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

\[13 + 13q^{2} - 23q =\]

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

\[9q² - 30q + 9 = 0\ \ \ \ |\ :3\]

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

\[D = 100 - 36 = 64\]

\[q_{1} = \frac{10 + 8}{6} = 3;\ \]

\[\ q_{2} = \frac{10 - 8}{6} = \frac{1}{3}\]

\[b_{1} \cdot \left( 1 + q^{2} - 2q \right) = 20\ \]

\[b_{1}(1 - q)^{2} = 20\ \ \]

\[b_{1} = \frac{20}{(1 - q)^{2}}\]

\[при\ \ q = 3:\ \ \]

\[b_{1} = \frac{20}{(1 - 3)^{2}} = \frac{20}{4} = 5,\]

\[\text{\ \ }b_{2} = 5 \cdot 3 = 15,\ \ \]

\[b_{3} = 15 \cdot 3 = 45;\]

\[при\ q = \frac{1}{3}:\ \ \]

\[b_{1} = \frac{20}{\left( 1 - \frac{1}{3} \right)^{2}} = \frac{20 \cdot 9}{4} = 45,\]

\[\text{\ \ }b_{2} = 45 \cdot \frac{1}{3} = 15,\]

\[\text{\ \ }b_{3} = 15 \cdot \frac{1}{3} = 5\]

\[Ответ:5,\ 15,\ 45\ \ \ или\ \ \ 45,\ 15,\ 5.\]