Решебник по алгебре 9 класс Мерзляк Задание 1039

\[\boxed{\mathbf{1039\ (1039).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

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

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

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

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

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

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

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

\[D = 169 - 144 = 25\]

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

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

\[при\ \ q = 1,5:\ \ \]

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

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

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

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

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

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

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

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

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

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

\[Ответ:\ \ 2\frac{2}{3};4;6;9.\]