Решебник по математике 6 класс Виленкин Часть 1, 2 Часть 1. Параграф 2. Действия со смешанными числами Задание 416

\[\boxed{\mathbf{2.}\mathbf{416}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \left( 4\frac{3^{\backslash 3}}{4} - 3\frac{1}{12} \right) \cdot 4 =\]

\[= \left( 4\frac{9}{12} - 3\frac{1}{12} \right) \cdot 4 = 1\frac{8}{12} \cdot 4 =\]

\[= 1\frac{2}{3} \cdot 4 = \frac{5}{3} \cdot 4 = \frac{20}{3} = 6\frac{2}{3}\]

\[\textbf{б)}\ \left( 5\frac{14^{\backslash 2}}{19} - 5\frac{1}{38} \right) \cdot 38 =\]

\[= \left( 5\frac{28}{38} - 5\frac{1}{38} \right) \cdot 38 = \frac{27}{38} \cdot 38 =\]

\[= 27\]

\[\textbf{в)}\ 7\frac{4}{19} \cdot 6\frac{1}{4} + 4\frac{15}{19} \cdot 6\frac{1}{4} =\]

\[= 6\frac{1}{4} \cdot \left( 7\frac{4}{19} + 4\frac{15}{19} \right) =\]

\[= \left( 6 + \frac{1}{4} \right) \cdot 12 =\]

\[= 6 \cdot 12 + \frac{1}{4} \cdot 12 = 72 + 3 = 75\]

\[\textbf{г)}\ 3\frac{1}{14} \cdot 17\frac{7}{29} - 3\frac{1}{14} \cdot 3\frac{7}{29} =\]

\[= 3\frac{1}{14} \cdot \left( 17\frac{7}{29} - 3\frac{7}{29} \right) =\]

\[= \left( 3 + \frac{1}{14} \right) \cdot 14 =\]

\[= 3 \cdot 14 + \frac{1}{14} \cdot 14 = 42 + 1 =\]

\[= 43\]

\[\textbf{д)}\ \left( 1\frac{1^{\backslash 8}}{2} + 2\frac{1}{16} \right) \cdot 2\frac{10}{11} =\]

\[= \left( 1\frac{8}{16} + 2\frac{1}{16} \right) \cdot 2\frac{10}{11} =\]

\[= 3\frac{9}{16} \cdot \frac{32}{11} = \frac{57 \cdot 32}{16 \cdot 11} = \frac{114}{11} =\]

\[= 10\frac{4}{11}\]

\[\textbf{е)}\ 2\frac{2}{3} \cdot \left( 2\frac{1}{16} - 1\frac{7^{\backslash 2}}{8} \right) =\]

\[= 2\frac{2}{3} \cdot \left( 1\frac{17}{16} - 1\frac{14}{16} \right) = \frac{8}{3} \cdot \frac{3}{16} =\]

\[= \frac{1}{2} = 0,5\]