Решебник по алгебре 9 класс Макарычев Задание 625

\[\boxed{\text{625\ (}\text{с}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[x_{n} = x_{1} \cdot q^{n - 1};\]

\[\textbf{а)}\ x_{1} = 16;\ \ q = \frac{1}{2}:\]

\[x_{7} = x_{1} \cdot q^{6} = 16 \cdot \left( \frac{1}{2} \right)^{6} = 2^{4} \cdot 2^{- 6} = 2^{- 2} = \frac{1}{4}.\]

\[\textbf{б)}\ x_{1} = - 810;\ \ q = \frac{1}{3}:\]

\[x_{8} = x_{1} \cdot q^{7} = - 810 \cdot \left( \frac{1}{3} \right)^{7} = - 10 \cdot 3^{4} \cdot 3^{- 7} = - \frac{10}{27}.\]

\[\textbf{в)}\ x_{1} = \sqrt{2};\ \ q = - \sqrt{2}:\]

\[x_{10} = x_{1} \cdot q^{9} = \sqrt{2} \cdot \left( - \sqrt{2} \right)^{9} = - \left( \sqrt{2} \right)^{10} = - 2^{5} = - 32.\]

\[\textbf{г)}\ x_{1} = - 125;\ \ q = 0,2:\]

\[x_{6} = x_{1} \cdot q^{5} = - 125 \cdot (0,2)^{5} = - 5^{3} \cdot \frac{1}{5^{5}} = - 5^{3} \cdot 5^{- 5} = - 5^{- 2} = - \frac{1}{25}.\]

\[\boxed{\text{625.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[b_{7} = 0,012;\ \ q = 0,2:\]

\[b_{7} = b_{1} \cdot q^{6} = 0,012;\]

\[0,012 = b_{1} \cdot {0,2}^{6}\]

\[b_{1} = \frac{0,012}{{0,2}^{6}} = 187,5\]

\[\Longrightarrow b_{n} = 187,5 \cdot (0,2)^{n - 1}.\]