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МатематикаРешебник по алгебре 9 класс Мерзляк Задание 739
Задание 739
\[\boxed{\mathbf{739\ (739).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[a_{1}a_{4} = a_{2}^{2}\text{\ \ \ }\]
\[\ если\ \ d = 0 \Longrightarrow \text{\ \ }a_{1} \cdot\]
\[\cdot \left( a_{1} + 3d \right) =\]
\[= \left( a_{1} + d \right)^{2} \Longrightarrow \text{\ \ }a_{1}^{2} = a_{1}^{2}\]
\[если\ \ a_{1} = d \Longrightarrow \text{\ \ }a_{1}\left( a_{1} + 3d \right) =\]
\[= \left( a_{1} + d \right)^{2}\]
\[d \cdot (d + 3d) = (d + d)^{2}\text{\ \ }\]
\[d \cdot 4d = (2d)^{2}\text{\ \ }\]
\[4d^{2} = 4d^{2}\ \]
\[Ответ:если\ d = 0\ \ или\]
\[\text{\ \ }a_{1} = d.\]
\[\boxed{\mathbf{739.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\ \left\{ \begin{matrix} b_{3} = 4 \\ b_{5} = 2 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\left\{ \begin{matrix} b_{1}q^{2} = 4 \\ b_{1}q^{4} = 2 \\ \end{matrix} \right.\ \ \ \ \ |(\ :)\ \ \]
\[\frac{b_{1}q^{2}}{b_{1}q^{4}} = \frac{4}{2}\text{\ \ }\]
\[\frac{1}{q^{2}} = 2 \Longrightarrow \ q^{2} = \frac{1}{2}\]
\[\left\{ \begin{matrix} q = \pm \frac{1}{\sqrt{2}} \\ b_{1} = \frac{4}{q^{2}}\text{\ \ \ } \\ \end{matrix} \Longrightarrow \text{\ \ }b_{1} = 4\ :\frac{1}{2} = 8 \right.\ \]
\[при\ \ b_{1} = 8 \Longrightarrow \ q = \frac{1}{\sqrt{2}}\]
\[S = \frac{8}{1 - \frac{1}{\sqrt{2}}} = \frac{8}{\frac{\sqrt{2} - 1}{\sqrt{2}}} =\]
\[= \frac{8\sqrt{2}}{\sqrt{2} - 1} = \frac{8\sqrt{2} \cdot \left( \sqrt{2} + 1 \right)}{\left( \sqrt{2} - 1 \right)\left( \sqrt{2} + 1 \right)} =\]
\[= \frac{8\sqrt{2} \cdot \left( \sqrt{2} + 1 \right)}{2 - 1} =\]
\[= 8 \cdot 2 + 8\sqrt{2} = 16 + 8\sqrt{2};\]
\[при\ \ b_{1} = 8 \Longrightarrow \ \ q = - \frac{1}{\sqrt{2}}\]
\[S = \frac{8}{1 + \frac{1}{\sqrt{2}}} = \frac{8\sqrt{2}}{\sqrt{2} + 1} =\]
\[= 16 - 8\sqrt{2}.\]
\[Ответ:16 \pm 8\sqrt{2}.\]
\[2)\ \left\{ \begin{matrix} b_{1} + b_{3} = 20 \\ b_{2} + b_{4} = \frac{20}{3} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} b_{1} + b_{1}q^{2} = 20 \\ b_{1}q + b_{1}q^{3} = \frac{20}{3} \\ \end{matrix}\ \ \ \ \ |(\ : \right.\ )\ \ \]
\[\left\{ \begin{matrix} q = \frac{1}{3}\text{\ \ \ \ \ \ \ \ \ \ \ \ } \\ b_{1} = \frac{20}{1 + q^{2}} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} q = \frac{1}{3}\text{\ \ \ \ \ \ \ \ } \\ b_{1} = \frac{20}{1 + \frac{1}{9}} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} q = \frac{1}{3}\text{\ \ \ \ \ \ \ \ \ \ \ } \\ b_{1} = \frac{20 \cdot 9}{10}\ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} q = \frac{1}{3}\text{\ \ \ \ } \\ b_{1} = 18 \\ \end{matrix} \right.\ \]
\[S = \frac{18}{1 - \frac{1}{3}} = \frac{18}{\frac{2}{3}} = \frac{18 \cdot 3}{2} = 27\]
\[Ответ:27.\]
