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МатематикаГДЗ по алгебре 9 класс Макарычев Задание 589
Задание 589
\[\boxed{\text{589\ (589).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ c_{5} = 27;\ \ c_{27} = 60:\]
\[\left\{ \begin{matrix} c_{1} + 4d = 27\ \ \\ c_{1} + 26d = 60 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} 22d = 33\ \ \ \ \ \\ c_{1} = 27 - 4d \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} d = 1,5\ \\ c_{1} = 21. \\ \end{matrix} \right.\ \]
\[Ответ:d = 1,5;\ \ c_{1} = 21.\]
\[\textbf{б)}\ c_{20} = 0;\ \ c_{66} = - 92:\]
\[\left\{ \begin{matrix} c_{1} + 19d = 0\ \ \ \ \ \ \\ c_{1} + 65d = - 92 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} 46d = - 92 \\ c_{1} = - 19d\ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} d = - 2\ \\ c_{1} = 38. \\ \end{matrix} \right.\ \]
\[Ответ:d = - 2;\ \ c_{1} = 38.\]
\[\boxed{\text{589.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ b_{1} = 6;\ \ q = 2:\ \ \]
\[b_{2} = b_{1} \cdot q = 6 \cdot 2 = 12,\ \ \]
\[b_{3} = b_{2} \cdot q = 12 \cdot 2 = 24,\ \ \]
\[b_{4} = b_{3} \cdot q = 24 \cdot 2 = 48,\ \ \]
\[b_{5} = b_{4} \cdot q = 48 \cdot 2 = 96.\]
\[\textbf{б)}\ b_{1} = - 16;\ \ q = \frac{1}{2}:\]
\[b_{2} = b_{1} \cdot q = - 16 \cdot \frac{1}{2} = - 8,\ \ \]
\[b_{3} = b_{2} \cdot q = - 8 \cdot \frac{1}{2} = - 4,\ \ \]
\[b_{4} = b_{3} \cdot q = - 4 \cdot \frac{1}{2} = - 2,\ \ \]
\[b_{5} = b_{4} \cdot q = - 2 \cdot \frac{1}{2} = - 1\text{.\ }\]
\[\textbf{в)}\ \ b_{1} = - 24;\ \ q = - 1,5:\ \]
\[b_{2} = b_{1} \cdot q = - 24 \cdot ( - 1,5) = 36,\ \ \]
\[b_{3} = b_{2} \cdot q = 36 \cdot ( - 1,5) = - 54,\ \]
\[b_{4} = b_{3} \cdot q = - 54 \cdot ( - 1,5) = 81,\ \ \]
\[b_{5} = b_{4} \cdot q = 81 \cdot ( - 1,5) =\]
\[= - 121,5.\]
\[\textbf{г)}\ \ b_{1} = 0,4;\ \ \ q = \sqrt{2}:\ \]
\[b_{2} = b_{1} \cdot q = 0,4 \cdot \sqrt{2},\ \ \]
\[b_{3} = b_{2} \cdot q = 0,4 \cdot \sqrt{2} \cdot \sqrt{2} = 0,8,\ \ \]
\[b_{4} = b_{3} \cdot q = 0,8\sqrt{2},\ \ \]
\[b_{5} = b_{4} \cdot q = 0,8 \cdot \sqrt{2} \cdot \sqrt{2} = 1,6.\]
