Решебник по математике 6 класс Виленкин Часть 1, 2 Часть 1. Параграф 2. Действия со смешанными числами Задание 322

\[\boxed{\mathbf{2.}\mathbf{322}}\]

\[\textbf{а)}\left( 1\frac{1}{2} \right)^{3} - 2\frac{1}{3} \cdot 1\frac{1}{4} =\]

\[= \left( \frac{3}{2} \right)^{3} - \frac{7}{3} \cdot \frac{5}{4} = \frac{27^{\backslash 3}}{8} - \frac{35^{\backslash 2}}{12} =\]

\[= \frac{81}{24} - \frac{70}{24} = \frac{11}{24}\]

\[\textbf{б)}\ \frac{4}{9} \cdot \left( 3\frac{3}{14} \cdot 2\frac{4}{5} \right)^{2} =\]

\[= \frac{4}{9} \cdot \left( \frac{45 \cdot 14}{14 \cdot 5} \right)^{2} = \frac{4}{9} \cdot (9)^{2} =\]

\[= \frac{4}{9} \cdot 81 = 4 \cdot 9 = 36\]

\[\textbf{в)}\ \left( \left( \frac{2}{3} \right)^{3} + \frac{5}{9} \right) \cdot \frac{9}{11} =\]

\[= \left( \frac{8}{27} + \frac{5^{\backslash 3}}{9} \right) \cdot \frac{9}{11} =\]

\[= \left( \frac{8}{27} + \frac{15}{27} \right) \cdot \frac{9}{11} = \frac{23}{27} \cdot \frac{9}{11} =\]

\[= \frac{23}{33}\]

\[= \left( 1\frac{21}{14} - \frac{11}{14} \right) \cdot 1\frac{19}{36} =\]

\[= 1\frac{10}{14} \cdot \frac{55}{36} = 1\frac{5}{7} \cdot \frac{55}{36} =\]

\[= \frac{12}{7} \cdot \frac{55}{36} = \frac{55}{21} = 2\frac{13}{21}\]