Решебник по математике 6 класс Виленкин Часть 1, 2 Часть 1. Параграф 2. Действия со смешанными числами Задание 464

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Пояснение.

Решение.

\[\textbf{а)}\ 2\frac{1}{7} \cdot \left( 2\frac{1}{4}\ :3\frac{6}{7} \right) =\]

\[= \frac{15}{7} \cdot \left( \frac{9}{4}\ :\frac{27}{7} \right) = \frac{15 \cdot 9 \cdot 7}{7 \cdot 4 \cdot 27} =\]

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

\[\textbf{б)}\ \left( 1\frac{2}{9} + 1\frac{5}{9} \right) \cdot 1\frac{4}{5} = 2\frac{7}{9} \cdot 1\frac{4}{5} =\]

\[= \frac{25}{9} \cdot \frac{9}{5} = 5\]

\[\textbf{в)}\ \left( 7\frac{1}{3} - 5\frac{1}{6} \right)\ :3\frac{1}{3} =\]

\[= \left( 7\frac{2}{6} - 5\frac{1}{6} \right)\ :\frac{10}{3} = 2\frac{1}{6} \cdot \frac{3}{10} =\]

\[= \frac{13}{10} \cdot \frac{3}{10} = \frac{39}{100} = 0,39\]

\[\textbf{г)}\ \left( 2\frac{2}{15} - 1\frac{2}{5} \right) \cdot 6\frac{1}{4} =\]

\[= \left( 1\frac{17}{15} - 1\frac{6}{15} \right) \cdot \frac{25}{4} = \frac{11}{15} \cdot \frac{25}{4} =\]

\[= \frac{11 \cdot 5}{3 \cdot 4} = \frac{55}{12} = 4\frac{7}{12}\]

\[\textbf{д)}\ \left( 2\frac{2}{3} + 1\frac{5}{6} \right)\ :4\frac{1}{2} =\]

\[= \left( 2\frac{4}{6} + 1\frac{5}{6} \right)\ :\frac{9}{2} = 3\frac{9}{6} \cdot \frac{2}{9} =\]

\[= 4\frac{3}{6} \cdot \frac{2}{9} = 4\frac{1}{2} \cdot \frac{2}{9} = \frac{9}{2} \cdot \frac{2}{9} = 1\]

\[\textbf{е)}\ \left( 7\frac{1}{8} - 6\frac{3}{5} \right)\ :4\frac{1}{5} =\]

\[= \left( 6\frac{9}{8} - 6\frac{3}{5} \right)\ :\frac{21}{5} =\]

\[= \left( 6\frac{45}{40} - 6\frac{24}{40} \right) \cdot \frac{5}{21} =\]

\[= \frac{21}{40} \cdot \frac{5}{41} = \frac{1}{8}\]