Решебник по математике 6 класс Мерзляк ФГОС Задание 320

\[\boxed{\mathbf{320\ (321).}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[\frac{1}{10} + \frac{1}{11} + \frac{1}{12} + .. + \frac{1}{18} > \frac{1}{2}\]

\[\frac{1}{10} > \frac{1}{18};\ \ \frac{1}{11} > \frac{1}{18};\ \ \frac{1}{12} > \frac{1}{18};\ \]

\[\ \frac{1}{13} > \frac{1}{18};\ \frac{1}{14} > \frac{1}{18};\ \frac{1}{15} > \frac{1}{18};\ \]

\[\frac{1}{16} > \frac{1}{18};\ \ \frac{1}{17} > \frac{1}{18}\]

\[\frac{1}{1} + \frac{1}{1} + .. + \frac{1}{18} > \frac{1}{18} + \frac{1}{18} + .. + \frac{1}{18} =\]

\[= \frac{9}{18} = \frac{1}{2} \Longrightarrow верно,\ ч.\ т.д.\]

\[\boxed{\mathbf{320\ (}\mathbf{с}\mathbf{).}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[\frac{2}{3 \cdot 5} + \frac{2}{5 \cdot 7} + .. + \frac{2}{29 \cdot 31} =\]

\[= \frac{1^{\backslash 31}}{3} - \frac{1^{\backslash 3}}{31} = \frac{31}{93} - \frac{3}{93} =\]

\[= \frac{31 - 3}{93} = \frac{28}{93}\]

\[\boxed{\mathbf{320}\mathbf{.}}\]

\[1)\ \frac{7}{4} = 1\frac{3}{4}\]

\[2)\ \frac{18}{11} = 1\frac{7}{11}\]

\[3)\ \frac{45}{7} = 6\frac{3}{7}\]

\[4)\ \frac{93}{23} = 4\frac{1}{23}\]

\[5)\ \frac{100}{17} = 5\frac{15}{17}\]