Решебник по алгебре 9 класс Макарычев Задание 568

\[\boxed{\text{568}\text{\ (568)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ b_{n + 1} = b_{n} + 3;\ \ \ b_{1} = 10:\ \]

\[b_{2} = b_{1} + 3 = 10 + 3 = 13\]

\[b_{3} = b_{2} + 3 = 13 + 3 = 16\]

\[b_{4} = b_{3} + 3 = 16 + 3 = 19\]

\[b_{5} = b_{4} + 3 = 19 + 3 = 22.\]

\[\textbf{б)}\ b_{n + 1} = \frac{b_{n}}{2};\ \ \ b_{1} = 40:\ \]

\[b_{2} = \frac{b_{1}}{2} = \frac{40}{2} = 20\]

\[b_{3} = \frac{b_{2}}{2} = \frac{20}{2} = 10\]

\[b_{4} = \frac{b_{3}}{2} = 5\]

\[b_{5} = \frac{b_{4}}{2} = 2,5.\]

\[\boxed{\text{568.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ 125^{- 1} \cdot 25² = 5^{- 3} \cdot 5^{4} = 5;\]

\[\textbf{б)}\ 0,0001 \cdot \left( 10^{3} \right)^{2} \cdot (0,1)^{- 2} =\]

\[= 10^{- 4} \cdot 10^{6} \cdot 10^{2} = 10^{4} =\]

\[= 10\ 000;\]

\[\textbf{в)}\ \frac{16^{- 3} \cdot 4^{5}}{8} = 2^{- 12} \cdot 2^{10} \cdot 2^{- 3} =\]

\[= 2^{- 5} = \frac{1}{32};\]

\[\textbf{г)}\ 9^{4} \cdot \left( \frac{1}{27} \right)^{- 3} \cdot 81^{- 4} =\]

\[= 3^{8} \cdot 3^{9} \cdot 3^{- 16} = 3.\ \]