Решебник по математике 6 класс Виленкин Часть 1, 2 Часть 1. Параграф 2. Действия со смешанными числами Задание 493

\[\boxed{\mathbf{2.493}}\]

\[\textbf{а)}\ \frac{5}{4}\ :\frac{2}{15} = \frac{5}{4} \cdot \frac{15}{2} = \frac{75}{8} = 9\frac{3}{8}\]

\[\textbf{б)}\ \frac{2}{3}\ :\frac{8}{9} = \frac{2}{3} \cdot \frac{9}{8} = \frac{3}{4}\]

\[\textbf{в)}\ \frac{64}{131}\ :\frac{32}{52} = \frac{64}{131} \cdot \frac{52}{32} = \frac{104}{131}\]

\[\textbf{г)}\ \frac{64}{125}\ :4 = \frac{64}{125} \cdot \frac{1}{4} = \frac{16}{125}\]

\[\textbf{д)}\ 9\ :\frac{3}{4} = 9 \cdot \frac{4}{3} = 3 \cdot 4 = 12\]

\[\textbf{е)}\ 9\ :4 = \frac{9}{4} = 2\frac{1}{4}\]

\[\textbf{ж)}\ 1\frac{3}{7}\ :1\frac{1}{14} = \frac{10}{7}\ :\frac{15}{14} =\]

\[= \frac{10}{7} \cdot \frac{14}{15} = \frac{2 \cdot 2}{3} = \frac{4}{3} = 1\frac{1}{3}\]

\[\textbf{з)}\ 2\frac{1}{3}\ :7\frac{1}{9} = \frac{7}{3}\ :\frac{64}{9} = \frac{7}{3} \cdot \frac{9}{64} =\]

\[= \frac{21}{64}\]

\[\textbf{и)}\ 3\frac{8}{19}\ :5\frac{15}{38} = \frac{65}{19}\ :\frac{205}{38} =\]

\[= \frac{65}{19} \cdot \frac{38}{205} = \frac{13 \cdot 2}{41} = \frac{26}{41}\]

\[к)\ 4\frac{17}{36}\ :19\frac{1}{6} = \frac{161}{36}\ :\frac{115}{6} =\]

\[= \frac{161}{36} \cdot \frac{6}{115} = \frac{7 \cdot 1}{6 \cdot 5} = \frac{7}{30}\ \]