Решебник по математике 6 класс Мерзляк ФГОС Задание 421

 

\[\boxed{\mathbf{421\ }\left( \mathbf{421} \right)\mathbf{.}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

Пояснение.

Решение.

\[a \neq 0;\ \ \ b \neq 0\]

\[1)\ \frac{3}{4}a = \frac{2}{3}\text{b\ \ \ \ }\]

\[a = \frac{2}{3}b\ :\frac{3}{4}\]

\[a = \frac{2}{3} \cdot \frac{4}{3} \cdot b\]

\[a = \frac{2 \cdot 4}{3 \cdot 3}\text{b\ \ \ }\]

\[a = \frac{8}{9}b\ \Longrightarrow a < b.\]

\[2)\ \frac{2}{5}a = \frac{5}{7}\text{b\ \ \ \ }\]

\[a = \frac{5}{7}b\ :\frac{2}{5}\]

\[a = \frac{5}{7} \cdot \frac{5}{2} \cdot b\]

\[a = \frac{5 \cdot 5}{7 \cdot 2}\text{b\ \ \ \ }\]

\[a = \frac{25}{14}b\ \Longrightarrow a > b.\]

\[\boxed{\mathbf{421}.}\]

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

\[\frac{1}{2} = \frac{9}{18}\]

\[\frac{9}{18} = \frac{1}{18} + \frac{1}{18} + \frac{1}{18} + \ldots + \frac{1}{18}\]

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

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

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

\[\frac{1}{10} + \frac{1}{11} + \frac{1}{12} + \ldots + \frac{1}{18} > \frac{9}{18}\]

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

\[Что\ и\ требовалось\ доказать.\]