Решебник по алгебре 9 класс Макарычев Задание 574

\[\boxed{\text{574}\text{\ (574)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 81 \cdot 3^{- 6} = 3^{4} \cdot 3^{- 6} = 3^{- 2} = \frac{1}{9}.\]

\[\textbf{б)}\ \frac{\left( - 3^{- 3} \right)^{3}}{- 9^{- 2}} = \frac{3^{- 9}}{3^{- 4}} = 3^{- 5} = \frac{1}{243}.\]

\[\textbf{в)}\ 9^{- 5} \cdot \left( \frac{1}{9} \right)^{- 3} =\]

\[= 3^{- 10} \cdot \left( 3^{- 2} \right)^{- 3} = 3^{- 10} \cdot 3^{6} =\]

\[= 3^{- 4} = \frac{1}{81}.\]

\[\textbf{г)}\ \left( - 3^{- 3} \right)^{2} \cdot 27³ = 3^{- 6} \cdot 3^{9} = 27.\ \]

\[\boxed{\text{574.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ x_{1} = 2,\ x_{2} = 4,\ldots,\ a_{n} = 2n,\ \]

\[S_{n} = \frac{\left( x_{1} + x_{n} \right)}{2} \cdot n = \frac{2 + 2n}{2} \cdot n =\]

\[= n(n + 1) = n^{2} + n.\]

\[\textbf{б)}\ x_{1} = 1,\ x_{2} = 3,\ \ldots,\ x_{n} = 2n - 1,\]

\[S_{n} = \frac{\left( x_{1} + x_{n} \right)}{2} \cdot n =\]

\[= \frac{1 + 2n - 1}{2} \cdot n = n^{2}.\]

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