Решебник по математике 6 класс Виленкин ФГОС Часть 1, 2 Часть 2. П. 4. Действия с рациональными числами Задание 290

\[\boxed{\text{4.\ 290}}\]

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

\[\frac{1}{3} - \frac{1}{3} = 0\]

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

\[3 - \frac{1}{3} = 2\frac{2}{3}\]

\[- \frac{1}{9} - \frac{1}{3} = - \frac{1}{9} - \frac{3}{9} = - \frac{4}{9}\]

\[\frac{1}{6} - \frac{1}{3} = \frac{1}{6} - \frac{2}{6} = - \frac{1}{6}\]

\[- \frac{2}{3} - \frac{1}{3} = - \frac{3}{3} = - 1\]

\[2\frac{5}{6} - \frac{1}{3} = 2\frac{5}{6} - \frac{2}{6} = 2\frac{3}{6} = 2\frac{1}{2}\]

\[\textbf{б)}\ 5 \cdot \frac{2}{5} = 2\]

\[\frac{5}{18} \cdot \frac{2}{5} = \frac{1}{9}\]

\[1 \cdot \frac{2}{5} = \frac{2}{5}\]

\[\frac{1}{6} \cdot \frac{2}{5} = \frac{1}{15}\]

\[0 \cdot \frac{2}{5} = 0\]

\[20 \cdot \frac{2}{5} = 8\]

\[\frac{1}{3} \cdot \frac{2}{5} = \frac{2}{15}\]

\[1\frac{2}{3} \cdot \frac{2}{5} = \frac{5}{3} \cdot \frac{2}{5} = \frac{2}{3}\]