Решебник по математике 6 класс Виленкин Часть 1, 2 Часть 2. Параграф 4. Действия с рациональными числами Задание 319

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\[\textbf{а)}\ 1\ :\frac{1}{7} = 1 \cdot 7 = 7\]

\[\frac{6}{21}\ :\frac{1}{7} = \frac{6}{21} \cdot 7 = \frac{6}{3} = 2\]

\[\frac{1}{9}\ :\frac{1}{7} = \frac{1}{9} \cdot 7 = \frac{7}{9}\]

\[\frac{5}{7}\ :\frac{1}{7} = \frac{5}{7} \cdot 7 = 5\]

\[2\ :\frac{1}{7} = 2 \cdot 7 = 14\]

\[\frac{1}{7}\ :\frac{1}{7} = 1\]

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

\[1\frac{2}{7}\ :\frac{1}{7} = \frac{9}{7}\ :\frac{1}{7} = \frac{9}{7} \cdot 7 = 9\]

\[\textbf{б)} - \frac{1}{3} + \frac{1}{12} = - \frac{4}{12} + \frac{1}{12} = - \frac{3}{12}\]

\[- 1 + \frac{1}{12} = - \frac{12}{12} + \frac{1}{12} = - \frac{11}{12}\]

\[\frac{5}{12} + \frac{1}{12} = \frac{6}{12} = \frac{1}{2}\]

\[- \frac{1}{4} + \frac{1}{12} = - \frac{3}{12} + \frac{1}{12} = - \frac{2}{12} =\]

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

\[\frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}\]

\[- \frac{1}{6} + \frac{1}{12} = - \frac{2}{12} + \frac{1}{12} = - \frac{1}{12}\]

\[- \frac{7}{12} + \frac{1}{12} = - \frac{6}{12} = - \frac{1}{2}\]

\[\frac{1}{4} + \frac{1}{12} = \frac{3}{12} + \frac{1}{12} = \frac{4}{12} = \frac{1}{3}\]