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МатематикаРешебник по алгебре 9 класс Мерзляк Задание 745
Задание 745
\[\boxed{\mathbf{745\ (745).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[Пусть\ прогрессия:a_{1};\ a_{1} + d;\]
\[\ a_{1} + 2d;\ a_{1} + 3d;\ a_{1} + 4d;\ \]
\[a_{1} + 5d;\ldots\]
\[Тогда\ новая:\ a_{1} + d;\ \ a_{1} + 3d;\ \]
\[\ a_{1} + 5d;\ \ldots\]
\[\mathbf{Тогда\ разность\ \ \ }a_{1} +\]
\[+ 3d{- (a}_{1} + d) = a_{1} + 3d - a_{1} -\]
\[- d = 2d \Longrightarrow образуют.\]
\[\mathbf{Ответ:да.}\]
\[\boxed{\mathbf{745.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[\left\{ \begin{matrix} c_{3}c_{5} = 20\ \ \ \ \ \ \ \ \ \ \ \ \ \\ c_{2} + c_{4} = 12\sqrt{5}\ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} c_{1}q^{2} \cdot c_{1}q^{4} = 20\ \ \ \ \ \\ c_{1}q + c_{1}q^{3} = 12\sqrt{5} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} c_{1}^{2} \cdot q^{6} = 20\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ c_{1}q\left( 1 + q^{2} \right) = 12\sqrt{5}\text{\ \ \ } \\ \end{matrix} \right.\ \text{\ \ }\]
\[\left\{ \begin{matrix} c_{1}q^{3} = \pm 2\sqrt{5}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ } \\ c_{1}q\left( 1 + q^{2} \right) = 12\sqrt{5} \\ \end{matrix} \right.\ \ \]
\[\left\{ \begin{matrix} c_{1}q^{3} = 2\sqrt{5}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ c_{1}q\left( 1 + q^{2} \right) = 12\sqrt{5} \\ \end{matrix} \right.\ \ \ |\ \ (:)\text{\ \ \ \ \ }\]
\[\ \frac{c_{1}q^{3}}{c_{1}q\left( 1 + q^{2} \right)} = \frac{2\sqrt{5}}{12\sqrt{5}}\text{\ \ }\]
\[\frac{q^{2}}{1 + q^{2}} = \frac{1}{6}\]
\[6q^{2} = 1 + q^{2}\]
\[5q^{2} = 1\]
\[q² = \frac{1}{5} \Longrightarrow \ \ q = \pm \sqrt{\frac{1}{5}}\]
\[c_{1} = \frac{2\sqrt{5}}{q^{3}}\ \]
\[c_{1} = \frac{2\sqrt{5}}{\pm \left( \sqrt{\frac{1}{5}} \right)^{3}} =\]
\[= \frac{2\sqrt{5} \cdot 5\sqrt{5}}{1} = \pm 50\]
\[При\ q = \sqrt{\frac{1}{5}} \Longrightarrow \ \ c_{1} = 50:\]
\[S = \frac{50}{1 - \frac{1}{\sqrt{5}}} = \frac{50}{\frac{\sqrt{5} - 1}{\sqrt{5}}} =\]
\[= \frac{50\sqrt{5}}{\sqrt{5} - 1} = \frac{50\sqrt{5} \cdot \left( \sqrt{5} + 1 \right)}{\left( \sqrt{5} - 1 \right)\left( \sqrt{5} + 1 \right)} =\]
\[= \frac{50\sqrt{5} \cdot \left( \sqrt{5} + 1 \right)}{5 - 1} =\]
\[= \frac{25 \cdot \left( 5 + \sqrt{5} \right)}{2}.\]
\[При\ q = - \sqrt{\frac{1}{5}} \Longrightarrow \ \ c_{1} = - 50:\]
\[S = \frac{- 50}{1 + \frac{1}{\sqrt{5}}} = \frac{- 50}{\frac{\sqrt{5} + 1}{\sqrt{5}}} =\]
\[= \frac{- 50\sqrt{5}}{\sqrt{5} + 1} = \frac{- 50\sqrt{5} \cdot \left( \sqrt{5} - 1 \right)}{\left( \sqrt{5} + 1 \right)\left( \sqrt{5} - 1 \right)} =\]
\[= \frac{- 50\sqrt{5} \cdot \left( \sqrt{5} - 1 \right)}{5 - 1} =\]
\[= - 25\sqrt{5} \cdot \left( \sqrt{5} - 1 \right) =\]
\[= \frac{- 25 \cdot \left( 5 - \sqrt{5} \right)}{2} =\]
\[= \frac{25 \cdot (\sqrt{5} - 5)}{2}.\]
\[Если\ c_{1}q^{3} =\]
\[= - 2\sqrt{5} \Longrightarrow \ \frac{q^{2}}{1 + q^{2}} = - \frac{1}{6}\]
\[- 6q^{2} = 1 + q^{2}\text{\ \ }\]
\[7q^{2} = - 1\ \ \]
\[q^{2} = - \frac{1}{7} < 0 \Longrightarrow нет\ решений.\]
\[Ответ:\ \ S = \frac{25 \cdot \left( \sqrt{5} - 5 \right)}{2};\ \ \]
\[\ S = \frac{25 \cdot \left( 5 + \sqrt{5} \right)}{2}.\ \]
