Решебник по математике 6 класс Виленкин ФГОС Часть 1, 2 Часть 1. П. 2. Действия со смешанными числами Задание 276

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Пояснение.

Решение.

\[\textbf{а)}\ \left( \frac{4^{\backslash 4}}{9} + \frac{5^{\backslash 3}}{12} \right) \cdot \frac{18}{31} =\]

\[= \left( \frac{16}{36} + \frac{15}{36} \right) \cdot \frac{18}{31} = \frac{31}{36} \cdot \frac{18}{31} = \frac{1}{2}\]

\[\textbf{б)}\ \frac{6}{25} \cdot \left( \frac{11^{\backslash 4}}{15} - \frac{9^{\backslash 3}}{20} \right) =\]

\[= \frac{6}{25} \cdot \left( \frac{44}{60} - \frac{27}{60} \right) = \frac{6}{25} \cdot \frac{17}{60} =\]

\[= \frac{6 \cdot 17}{25 \cdot 60} = \frac{17}{25 \cdot 10} = \frac{17}{250}\]

\[\textbf{в)}\ \left( 4 - 3\frac{7}{15} \right) \cdot \frac{5}{8} = \frac{8}{15} \cdot \frac{5}{8} =\]

\[= \frac{8 \cdot 5}{15 \cdot 8} = \frac{1}{3}\]

\[\textbf{г)}\ \left( 5 - 4\frac{4}{7} \right) \cdot \left( 7\frac{1^{\backslash 2}}{6} - 6\frac{5}{12} \right) =\]

\[= \frac{3}{7} \cdot \left( 7\frac{2}{12} - 6\frac{5}{12} \right) =\]

\[= \frac{3}{7} \cdot \left( 6\frac{14}{12} - 6\frac{5}{12} \right) =\]

\[= \frac{3}{7} \cdot \frac{9}{12} = \frac{3 \cdot 9}{7 \cdot 12} = \frac{9}{7 \cdot 4} = \frac{9}{28}\]

\[\textbf{д)}\ \left( 1\frac{1}{24} - \frac{5^{\backslash 2}}{12} \right) \cdot \left( 4\frac{1^{\backslash 3}}{8} - 3\frac{5}{24} \right) =\]

\[= \left( \frac{25}{24} - \frac{10}{24} \right) \cdot \left( 4\frac{3}{24} - 3\frac{5}{24} \right) =\]

\[= \frac{15}{24} \cdot \left( 3\frac{27}{24} - 3\frac{5}{24} \right) = \frac{15}{24} \cdot \frac{22}{24} =\]

\[= \frac{15 \cdot 22}{24 \cdot 24} = \frac{5 \cdot 11}{12 \cdot 8} = \frac{55}{96}\]

\[\textbf{е)}\ \left( 1\frac{2}{15} - \frac{11}{15} \right) \cdot \left( 5\frac{3^{\backslash 3}}{18} - 4\frac{1^{\backslash 2}}{27} \right) =\]

\[= \left( \frac{17}{15} - \frac{11}{15} \right) \cdot \left( 5\frac{9}{54} - 4\frac{2}{54} \right) =\]

\[= \frac{6}{15} \cdot 1\frac{7}{54} =\]

\[= \frac{2}{5} \cdot \frac{61}{54} = \frac{2 \cdot 61}{5 \cdot 54} = \frac{61}{5 \cdot 27} = \frac{61}{135}\]