Решебник по математике 5 класс Ткачева рабочая тетрадь | Страница 68

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

\[1)\frac{13}{15} \cdot \frac{7}{8} \cdot \frac{8}{7} = \frac{13}{15} \cdot \left( \frac{7}{8} \cdot \frac{8}{7} \right) =\]

\[= \frac{13}{15} \cdot 1 = \frac{13}{15}\]

\[2)\ \frac{4}{9} \cdot \frac{6}{11} \cdot 2\frac{1}{4} = \frac{4}{9} \cdot \frac{6}{11} \cdot \frac{9}{4} =\]

\[= \left( \frac{4}{9} \cdot \frac{9}{4} \right) \cdot \frac{6}{11} = 1 \cdot \frac{6}{11} = \frac{6}{11}\]

\[3)\ \frac{3}{20} \cdot 1\frac{5}{6} \cdot \frac{6}{11} = \frac{3}{20} \cdot \frac{11}{6} \cdot \frac{6}{11} =\]

\[= \frac{3}{20} \cdot 1 = \frac{3}{20}\]

\[\boxed{\mathbf{3.}}\]

\[1)\ \frac{8}{15}\ :\frac{6}{35} = \frac{8}{15} \cdot \frac{35}{6} = \frac{8 \cdot 35}{15 \cdot 6} =\]

\[= \frac{4 \cdot 7}{3 \cdot 3} = \frac{28}{9} = 3\frac{1}{9}\]

\[2)\ \frac{10}{11}\ :25 = \frac{10}{11} \cdot \frac{1}{25} = \frac{10}{11 \cdot 25} =\]

\[= \frac{2}{11 \cdot 5} = \frac{2}{55}\]

\[3)\ \ 6\ :\frac{3}{4} = 6 \cdot \frac{4}{3} = \frac{6 \cdot 4}{3} = 2 \cdot 4 =\]

\[= 8\]

\[4)\ 1\frac{2}{7}\ \ :\frac{12}{21} = \frac{9}{7} \cdot \frac{21}{12} = \frac{9 \cdot 21}{7 \cdot 12} =\]

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

\[5)\ 2\frac{7}{10}\ :1\frac{2}{25} = \frac{27}{10}\ :\frac{27}{25} =\]

\[= \frac{27 \cdot 25}{10 \cdot 27} = \frac{5}{2} = 2\frac{1}{2}\]

\[\boxed{\mathbf{4.}}\]

\[1)\ 60\ :\frac{5}{12} = \frac{60}{1} \cdot \frac{12}{5} = \frac{60 \cdot 12}{5} =\]

\[= 12 \cdot 12 = 24.\]

\[2)\ 42\ :\frac{3}{7} = \frac{42}{1} \cdot \frac{7}{3} = \frac{42 \cdot 7}{3} =\]

\[= 14 \cdot 7 = 98.\ \ \]

\[\mathbf{41.\ 1.\ Основное\ свойство\ пропорции}\]

\[\boxed{\mathbf{1.}}\]

\[1)\ 2\ :3 = 4\ :6\]

\[\frac{15}{3} = \frac{20}{4}\]

\[1\frac{3}{5}\ :2 = \frac{4}{5}\]

\[2)\ \frac{42}{9} = \frac{14}{3}\]

\[150\ :20 = 37\frac{1}{2}\ :5\]

\[\frac{6}{11} = 18\ :33\]