ГДЗ по математике 6 класс Виленкин Часть 1, 2 Часть 1. Параграф 2. Действия со смешанными числами Задание 583

\[\boxed{\mathbf{2.583}}\]

\[\textbf{а)}\ \frac{2,16 \cdot 0,55 \cdot 4,5}{2,7 \cdot 0,15 \cdot 1,2} =\]

\[= \frac{1,8 \cdot 11 \cdot 5}{3 \cdot 3} = \frac{9 \cdot 11}{9} = 11\]

\[\textbf{б)}\ \frac{2\frac{2}{3} \cdot 2\frac{3}{7} \cdot \frac{9}{11}}{3\frac{3}{7} \cdot 5\frac{2}{3} \cdot \frac{1}{11}} = \frac{\frac{8}{3} \cdot \frac{17}{7} \cdot \frac{9}{11}}{\frac{24}{7} \cdot \frac{17}{3} \cdot \frac{1}{11}} =\]

\[= \frac{8 \cdot 17 \cdot 9}{3 \cdot 7 \cdot 11} \cdot \frac{7 \cdot 3 \cdot 11}{24 \cdot 17 \cdot 1} = \frac{9}{3} = 3\]

\[\textbf{в)}\ \frac{11\frac{1}{4}\ :\frac{9}{32}}{21\frac{1}{3} \cdot 5\frac{1}{4} - 10\frac{2}{13} \cdot 4\frac{8}{11}} =\]

\[= \frac{40}{64} = \frac{5}{8}\]

\[\frac{45}{4} \cdot \frac{32}{9} = 5 \cdot 8 = 40\]

\[\frac{64}{3} \cdot \frac{21}{4} = 16 \cdot 7 = 112\]

\[\frac{132}{13} \cdot \frac{52}{11} = 12 \cdot 4 = 48\]

\[112 - 48 = 64\]

\[\textbf{г)}\ \frac{30,6\ :14\frac{4}{7} + 13,2\ :1\frac{1}{3}}{1\frac{5}{16}\ :1,75} =\]

\[= 12\ :\frac{3}{4} = 12 \cdot \frac{4}{3} = 4 \cdot 4 = 16\]

\[\frac{306}{10}\ :\frac{102}{7} = \frac{306 \cdot 7}{10 \cdot 102} = \frac{3 \cdot 7}{10} =\]

\[= \frac{21}{10} = 2,1\]

\[\frac{132}{10}\ :\frac{4}{3} = \frac{132 \cdot 3}{10 \cdot 4} = \frac{33 \cdot 3}{10} =\]

\[= \frac{99}{10} = 9,9\]

\[2,1 + 9,9 = 12\]

\[\frac{21}{16}\ :\frac{175}{100} = \frac{21}{16}\ :\frac{7}{4} = \frac{21 \cdot 4}{16 \cdot 7} = \frac{3 \cdot 4}{16} = \frac{3}{4}\]