Решебник по математике 6 класс Мерзляк ФГОС Задание 333

\[\boxed{\mathbf{333\ (333).}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{2}{13} \cdot 5 = \frac{2 \cdot 5}{13} = \frac{10}{13}\]

\[2)\ \frac{4}{17} \cdot 3 = \frac{4 \cdot 3}{17} = \frac{12}{17}\]

\[3)\ \frac{8}{9} \cdot 2 = \frac{8 \cdot 2}{9} = \frac{16}{9} = 1\frac{7}{9}\]

\[4)\frac{4}{49} \cdot 7 = \frac{4 \cdot 7}{49} = \frac{4}{7}\]

\[5)\ 7 \cdot \frac{3}{40} = \frac{7 \cdot 3}{40} = \frac{21}{40}\]

\[6)\ 6 \cdot \frac{15}{18} = \frac{6 \cdot 15}{18} = \frac{1 \cdot 15}{3} = 5\]

\[7)\frac{7}{12} \cdot 24 = \frac{7 \cdot 24}{12} = 7 \cdot 2 = 14\]

\[8)\ 45 \cdot \frac{8}{15} = \frac{45 \cdot 8}{15} = 3 \cdot 8 = 24\]

\[\boxed{\mathbf{333}\mathbf{.}}\]

\[1)\ \frac{16}{23} > \frac{9}{23}\]

\[2)\ \frac{3}{98} < \frac{3}{94}\]

\[3)\ 1 < \frac{22}{19}\]

\[4)\ \frac{3}{8} < \frac{1}{2} = \frac{4}{8}\]

\[5)\ \frac{5}{7} = \frac{15}{21} > \frac{2}{3} = \frac{14}{21}\]

\[6)\ \frac{5}{16} = \frac{15}{48} > \frac{7}{24} = \frac{14}{48}\]

\[7)\ \frac{8}{15} = \frac{32}{60} < \frac{11}{20} = \frac{33}{60}\]

\[8)\ \frac{13}{18} = \frac{39}{54} > \frac{19}{27} = \frac{38}{54}\]