Решебник по алгебре 9 класс Мерзляк Задание 698

\[\boxed{\mathbf{698\ (698).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\text{\ a}_{1} = 4,\ \ a_{n + 1} = a_{n} + 3\]

\[a_{2} = 4 + 3 = 7\]

\[a_{3} = 7 + 3 = 10\]

\[a_{4} = 10 + 3 = 13\]

\[a_{5} = 13 + 3 = 16\]

\[2)\ a_{1} = - 2,\ \ a_{2} = 6,\ \ \]

\[a_{n + 2} = 3a_{n} + a_{n} + 1\]

\[a_{3} = 3 \cdot ( - 2) + 6 = 0\]

\[a_{4} = 3 \cdot 6 + 0 = 18\]

\[a_{5} = 3 \cdot 0 + 18 = 18\]

\[\boxed{\mathbf{698.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{7^{9}}{7^{10}} = \frac{1}{7}\]

\[2)\ \frac{125^{3}}{25^{4}} = \frac{\left( 5^{3} \right)^{3}}{\left( 5^{2} \right)^{4}} = \frac{5^{9}}{5^{8}} = 5^{1} = 5\]

\[3)\ \frac{32^{5}}{64^{4}} = \frac{\left( 2^{5} \right)^{5}}{\left( 2^{6} \right)^{4}} = \frac{2^{25}}{2^{24}} = 2^{1} = 2\]

\[4)\ \frac{39^{8}}{3^{10} \cdot 13^{7}} = \frac{3^{8} \cdot 13^{8}}{3^{10} \cdot 13^{7}} =\]

\[= \frac{13}{3^{2}} = \frac{13}{9}\]