Решебник по математике 6 класс Виленкин ФГОС Часть 1, 2 Часть 1. П. 3. Отношения и пропорции Задание 140

\[\boxed{\mathbf{3.140}}\]

\[\textbf{а)}\ \frac{3}{5}?\frac{1}{6} = \frac{23}{30}\]

\[\frac{3}{5} + \frac{1}{6} = \frac{18}{30} + \frac{5}{30} = \frac{23}{30}\]

\[\textbf{б)}\ \frac{17}{36}?\frac{4}{9} = \frac{17}{81}\]

\[\frac{17}{36} \cdot \frac{4}{9} = \frac{17}{81}\]

\[\textbf{в)}\ \frac{9}{10}?1\frac{19}{45} = 1,28\]

\[\frac{9}{10} \cdot \frac{64}{45} = \frac{64}{10 \cdot 5} = \frac{64}{50} = \frac{128}{100} =\]

\[= 1,28\]

\[\textbf{г)}\ 2\frac{2}{5}?1\frac{3}{5} = 3\frac{21}{25}\]

\[2\frac{2}{5} \cdot 1\frac{3}{5} = \frac{12}{5} \cdot \frac{8}{5} = \frac{96}{25} = 3\frac{21}{25}\]

\[\textbf{д)}\ 3\frac{7}{9}?2\frac{35}{81} = 1\frac{28}{81}\]

\[3\frac{7}{9} - 2\frac{35}{81} = 3\frac{63}{81} - 2\frac{35}{81} = 1\frac{28}{81}\]

\[\textbf{е)}\ 1\frac{8}{9}?\frac{1}{6} = 11\frac{1}{3}\]

\[1\frac{8}{9}\ :\frac{1}{6} = \frac{17}{9} \cdot 6 = \frac{34}{3} = 11\frac{1}{3}\]