Решебник по алгебре 9 класс Мерзляк Задание 909

\[\boxed{\mathbf{909.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[S = 2;\ \ S_{4} = 1\frac{7}{8}\]

\[S = \frac{b_{1}}{1 - q} = 2 \Longrightarrow \ \ b_{1} =\]

\[= 2 \cdot (1 - q)\]

\[S_{4} = \frac{b_{1}\left( q^{4} - 1 \right)}{q - 1} = 1\frac{7}{8}\text{\ \ \ }\]

\[\ \frac{b_{1}(q^{2} - 1)(q^{2} + 1)}{q - 1} = \frac{15}{8}\]

\[b_{1}(q + 1)\left( q^{2} + 1 \right) = \frac{15}{8}\]

\[\left\{ \begin{matrix} b_{1} = 2 \cdot (1 - q)\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ b_{1}(q + 1)\left( q^{2} + 1 \right) = \frac{15}{8} \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\text{\ \ \ }\left\{ \begin{matrix} b_{1} = 2 \cdot (1 - q)\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ b_{1} = \frac{15}{8 \cdot (q + 1)\left( q^{2} + 1 \right)} \\ \end{matrix} \right.\ \]

\[\frac{15}{8 \cdot (q + 1)\left( q^{2} + 1 \right)} = 2 \cdot (1 - q)\]

\[15 = 16 \cdot (q + 1)(1 - q)\left( q^{2} + 1 \right)\]

\[15 = 16 \cdot \left( 1 - q^{2} \right)\left( q^{2} + 1 \right)\]

\[15 = 16 \cdot \left( 1 - q^{4} \right)\]

\[1 - q^{4} = \frac{15}{16}\]

\[q^{4} = 1 - \frac{15}{16},\ \ q^{4} = \frac{1}{16},\]

\[\ \ q = \pm \frac{1}{2}\]

\[при\ \ q = \frac{1}{2} \Longrightarrow \ \ b_{1} =\]

\[= 2 \cdot (1 - q) =\]

\[= 2 \cdot \left( 1 - \frac{1}{2} \right) = 2 \cdot \frac{1}{2} = 1;\]

\[при\ q = - \frac{1}{2} \Longrightarrow \ \ b_{1} =\]

\[= 2 \cdot \left( 1 + \frac{1}{2} \right) = 2 \cdot \frac{3}{2} = 3.\]

\[Ответ:\ b_{1} = 1;\ q = \frac{1}{2}\ или\ \]

\[\text{\ \ }b_{1} = 3;\ \ q = - \frac{1}{2}.\]