Решебник по математике 6 класс Виленкин Часть 1, 2 Часть 1. Параграф 2. Действия со смешанными числами Задание 450

\[\boxed{\mathbf{2.450}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \frac{4}{5}\ :\frac{9}{11} = \frac{4}{5} \cdot \frac{11}{9} = \frac{44}{45}\]

\[\textbf{б)}\ \frac{1}{7}\ :\frac{5}{8} = \frac{1}{7} \cdot \frac{8}{5} = \frac{8}{35}\]

\[\textbf{в)}\ \frac{2}{7}\ :\frac{2}{9} = \frac{2}{7} \cdot \frac{9}{2} = \frac{9}{7} = 1\frac{2}{7}\]

\[\textbf{г)}\ \frac{5}{8}\ :\frac{7}{12} = \frac{5}{8} \cdot \frac{12}{7} = \frac{5 \cdot 3}{2 \cdot 7} =\]

\[= \frac{15}{14} = 1\frac{1}{14}\]

\[\textbf{д)}\ \frac{4}{7}\ :\frac{16}{49} = \frac{4}{7} \cdot \frac{49}{16} = \frac{7}{4} = 1\frac{3}{4}\]

\[\textbf{е)}\ \frac{5}{9}\ :\frac{1}{2} = \frac{5}{9} \cdot 2 = \frac{10}{9} = 1\frac{1}{9}\]

\[\textbf{ж)}\ \frac{12}{25}\ :\frac{8}{15} = \frac{12}{25} \cdot \frac{15}{8} = \frac{3 \cdot 3}{5 \cdot 2} =\]

\[= \frac{9}{10} = 0,9\]

\[\textbf{з)}\ \frac{9}{14}\ :\frac{18}{35} = \frac{9}{14} \cdot \frac{35}{18} = \frac{5}{2 \cdot 2} =\]

\[= \frac{5}{4} = 1\frac{1}{4}\]