Решебник по математике 5 класс Виленкин Часть 1, 2 Часть 2. Параграф 6. Десятичные дроби Задание 59

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\[\textbf{а)}\ 7\frac{3}{5} - 1\frac{2}{5} = 6\frac{1}{5};\ \]

\[\textbf{б)}\ 28\frac{6}{13} - 7\frac{1}{13} = 21\frac{5}{13};\ \]

\[\textbf{в)}\ 6 - \frac{2}{5} = 5\frac{5}{5} - \frac{2}{5} = 5\frac{3}{5};\ \]

\[\textbf{г)}\ 21 - \frac{5}{9} = 20\frac{9}{9} - \frac{5}{9} = 20\frac{4}{9};\ \]

\[\textbf{д)}\ 4\frac{1}{7} - 1\frac{4}{7} = 3\frac{8}{7} - 1\frac{4}{7} = 2\frac{4}{7};\ \]

\[\textbf{е)}\ 12\frac{2}{5} - 1\frac{4}{5} = 11\frac{7}{5} - 1\frac{4}{5} =\]

\[= 10\frac{3}{5};\ \]

\[\textbf{ж)}\ 4\frac{1}{2} - 3\frac{1}{4} = 4\frac{2}{4} - 3\frac{1}{4} = 1\frac{1}{4};\ \]

\[\textbf{з)}\ 5\frac{1}{7} - 3\frac{20}{21} = 5\frac{3}{21} - 3\frac{20}{21} =\]

\[= 4\frac{24}{21} - 3\frac{20}{21} = 1\frac{4}{21}\text{.\ \ }\]