Решебник по математике 6 класс Виленкин Часть 1, 2 Часть 1. Параграф 2. Действия со смешанными числами Задание 452

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Пояснение.

Решение.

\[\textbf{а)}\ 5\frac{1}{4}\ :\frac{4}{5} = \frac{21}{4} \cdot \frac{5}{4} = \frac{105}{16} = 6\frac{9}{16}\]

\[\textbf{б)}\ \frac{2}{13}\ :2\frac{2}{13} = \frac{2}{13}\ :\frac{28}{13} =\]

\[= \frac{2}{13} \cdot \frac{13}{28} = \frac{1}{14}\]

\[\textbf{в)}\ 2\frac{4}{7}\ :1\frac{3}{10} = \frac{18}{7}\ :\frac{13}{10} =\]

\[= \frac{18}{7} \cdot \frac{10}{13} = \frac{180}{91} = 1\frac{89}{91}\]

\[\textbf{г)}\ 10\frac{3}{5}\ :3\frac{3}{5} = \frac{53}{5}\ :\frac{18}{5} =\]

\[= \frac{53}{5} \cdot \frac{5}{18} = \frac{53}{18} = 2\frac{17}{18}\]

\[\textbf{д)}\ 5\frac{1}{4}\ :1\frac{3}{4} = \frac{21}{4}\ :\frac{7}{4} = \frac{21}{4} \cdot \frac{4}{7} = 3\]

\[\textbf{е)}\ 5\frac{3}{7}\ :3 = \frac{38}{7} \cdot \frac{1}{3} = \frac{38}{21} = 1\frac{17}{21}\]

\[\textbf{ж)}\ 0\ :10\frac{7}{15} = 0\]

\[\textbf{з)}\ 5\frac{1}{16}\ :1 = 5\frac{1}{16}\]