Решебник по алгебре 9 класс Макарычев Задание 672

 

\[\boxed{\text{672}\text{\ (672)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ y_{1} = - 3;\ \ \ \ \ \ y_{n + 1} = 10 + y_{n}\ \]

\[y_{2} = - 3 + 10 = 7,\ \ \]

\[y_{3} = 7 + 10 = 17,\]

\[y_{4} = 17 + 10 = 27,\ \ \]

\[y_{5} = 27 + 10 = 37.\]

\[\textbf{б)}\ y_{1} = 10;\ \ \ y_{n + 1} = \frac{2,5}{y_{n}}\]

\[y_{2} = 2,5\ :10 = 0,25,\]

\[y_{3} = 2,5\ :0,25 = 10,\]

\[y_{4} = 2,5\ :10 = 0,25,\]

\[y_{5} = 2,5\ :0,25 = 10.\]

\[\textbf{в)}\ y_{1} = 1,5;\ \ \ \ y_{n + 1} = n + y_{n}\]

\[y_{2} = 1,5 + 1 = 2,5;\]

\[y_{3} = 2,5 + 2 = 4,5\]

\[y_{4} = 4,5 + 3 = 7,5;\]

\[y_{5} = 7,5 + 4 = 11,5.\]

\[\textbf{г)}\ y_{1} = - 4;\ \ \ \ y_{n + 1} = - n^{2} \cdot y_{n}\]

\[y_{2} = - 1^{2} \cdot ( - 4) = 4,\]

\[y_{3} = - 2^{2} \cdot 4 = - 16,\]

\[y_{4} = - 3^{2} \cdot ( - 16) = 144,\]

\[y_{5} = - 4^{2} \cdot 144 = - 2304.\]

\[\boxed{\text{672.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ b_{1} = \frac{243}{256},\ \ q = \frac{2}{3},\]

\[\ \ n = 8;\]

\[b_{8} = b_{1} \cdot q^{7} = \frac{243}{256} \cdot \left( \frac{2}{3} \right)^{7} =\]

\[= \frac{3^{5}}{2^{8}} \cdot \frac{2^{7}}{3^{7}} = \frac{1}{18}\text{.\ }\]

\[\textbf{б)}\ b_{1} = \sqrt{\frac{2}{3}},\ \ q = - \sqrt{6},\]

\[\ \ n = 5;\]

\[b_{5} = b_{1} \cdot q^{4} =\]

\[= \sqrt{\frac{2}{3}} \cdot \left( - \sqrt{6} \right)^{4} =\]

\[= 36 \cdot \sqrt{\frac{2}{3}} = 12\sqrt{6}.\]