Решебник по математике 6 класс Мерзляк ФГОС Задание 358

\[\boxed{\mathbf{358\ (358).}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[\boxed{\mathbf{358}\mathbf{.}}\]

\[1)\ \frac{1}{9}\ :\frac{1}{8} = \frac{1}{9} \cdot 8 = \frac{8}{9}\]

\[2)\ \frac{5}{8}\ :\frac{15}{16} = \frac{5 \cdot 16}{8 \cdot 15} = \frac{2}{3}\]

\[3)\ \frac{15}{68}\ :\frac{10}{51} = \frac{15}{68} \cdot \frac{51}{10} = \frac{15 \cdot 51}{68 \cdot 10} =\]

\[= \frac{3 \cdot 3}{4 \cdot 2} = \frac{9}{8} = 1\frac{1}{8}\]

\[4)\ 6\ :\frac{7}{9} = \frac{6 \cdot 9}{7} = \frac{54}{7} = 7\frac{5}{7}\]

\[5)\ \frac{7}{9}\ :5 = \frac{7}{9 \cdot 5} = \frac{7}{45}\]

\[6)\ 7\frac{1}{3}\ :1\frac{2}{9} = \frac{22}{3}\ :\frac{11}{9} = \frac{22 \cdot 9}{3 \cdot 11} =\]

\[= 2 \cdot 3 = 6\]

\[7)\ 3\frac{1}{3}\ :2\frac{6}{7} = \frac{10}{3}\ :\frac{20}{7} = \frac{10 \cdot 7}{3 \cdot 20} =\]

\[= \frac{7}{6}\]

\[8)\ 2\frac{10}{13}\ :3\frac{3}{26} = \frac{36}{13}\ :\frac{81}{26} =\]

\[= \frac{36 \cdot 26}{13 \cdot 81} = \frac{4 \cdot 2}{1 \cdot 9} = \frac{8}{9}\]