ГДЗ по математике 6 класс Дорофеев Задание 927

\[\boxed{\mathbf{927\ (927)}\mathbf{.}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[\textbf{а)} - \frac{1}{3} \bullet 2 = - \frac{1 \bullet 2}{3} = - \frac{2}{3}\]

\[\textbf{б)}\frac{2}{5} \bullet ( - 7) = \frac{2}{5} \bullet \left( - \frac{7}{1} \right) = - \frac{2 \bullet 7}{5 \bullet 1} = - \frac{14}{5} = - 2\frac{4}{5}\]

\[\textbf{в)} - \frac{4}{9} \bullet \frac{3}{8} = - \frac{4 \bullet 3}{9 \bullet 8} = - \frac{1}{6}\]

\[\textbf{г)} - \frac{7}{8} \bullet \left( - \frac{2}{7} \right) = + \frac{7 \bullet 2}{8 \bullet 7} = + \frac{1}{4} = \frac{1}{4}\]

\[\textbf{д)} - 1\frac{1}{2} \bullet \frac{2}{3} = - \frac{3}{2} \bullet \frac{2}{3} = - \frac{3 \bullet 2}{2 \bullet 3} = - 1\]

\[\textbf{е)}\ 2\frac{1}{5} \bullet \left( - \frac{3}{22} \right) = \frac{11}{5} \bullet \left( - \frac{3}{22} \right) = - \frac{11 \bullet 3}{5 \bullet 22} = - \frac{3}{10}\]