Решебник по алгебре 9 класс Макарычев Задание 626

\[\boxed{\text{626\ (}\text{с}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ b_{1} = \frac{3}{4};\ \ q = \frac{2}{3}:\]

\[b_{5} = b_{1} \cdot q^{4} = \frac{3}{4} \cdot \left( \frac{2}{3} \right)^{4} = \frac{3}{2^{2}} \cdot \frac{2^{4}}{3^{4}} = \frac{2^{2}}{3^{3}} = \frac{4}{27}.\]

\[\textbf{б)}\ b_{1} = 1,8;\ \ \ q = \frac{\sqrt{3}}{3}:\]

\[b_{4} = b_{1} \cdot q^{3} = 1,8 \cdot \left( \frac{\sqrt{3}}{3} \right)^{3} = 1,8 \cdot \left( \frac{1}{\sqrt{3}} \right)^{3} = \frac{1,8}{3} \cdot \frac{1}{\sqrt{3}} = \frac{0,6}{\sqrt{3}} = \frac{\sqrt{3}}{5}.\]

\[\boxed{\text{626.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \frac{2^{n + 2} - 2^{n - 1}}{2^{n}} =\]

\[= \frac{2^{n} \cdot \left( 2^{2} - 2^{- 2} \right)}{2^{n}} = 2^{2} - \frac{1}{4} =\]

\[= 4 - \frac{1}{4} = 3\frac{3}{4};\]

\[\textbf{б)}\ \frac{25^{n} - 5^{2n - 1}}{5^{2n}} = \frac{5^{2n} - 5^{2n - 1}}{5^{2n}} =\]

\[= \frac{5^{2n} \cdot (1 - 5^{- 1})}{5^{2n}} = 1 - \frac{1}{5} = \frac{4}{5}.\]