Решебник по математике 5 класс Виленкин ФГОС Часть 1, 2 Часть 2. Параграф 6. Десятичные дроби Задание 102

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\[\textbf{а)}\ a = \frac{7}{13}:\ \]

\[\frac{7}{13} + \frac{4}{13} = \frac{11}{13};\ \]

\[\frac{11}{13} + \frac{1}{13} = \frac{12}{13};\ \]

\[\frac{12}{13} - \frac{8}{13} = \frac{4}{13} > 1 \rightarrow нет;\ \]

\[\frac{4}{13} = 1 \rightarrow нет;\ \]

\[\frac{4}{13} + 1\frac{3}{13} = 1\frac{7}{13};\ \]

\[1\frac{7}{13} + 2 = 3\frac{7}{13}\text{.\ }\]

\[\textbf{б)}\ a = \frac{16}{13}:\ \]

\[\frac{16}{13} + \frac{4}{13} = \frac{20}{13};\ \]

\[\frac{20}{13} + \frac{1}{13} = \frac{21}{13};\ \]

\[\frac{21}{13} - \frac{8}{13} = \frac{13}{13} > 1 \rightarrow нет;\ \]

\[\frac{13}{13} = 1 \rightarrow да;\ \]

\[1 - \frac{3}{10} = \frac{7}{10}.\]

\[\textbf{в)}\ a = \frac{4}{13}:\ \]

\[\frac{4}{13} + \frac{4}{13} = \frac{8}{13};\ \]

\[\frac{8}{13} + \frac{1}{13} = \frac{9}{13};\ \]

\[\frac{9}{13} - \frac{8}{13} = \frac{1}{13} > 1 \rightarrow нет;\ \]

\[\frac{1}{13} = 1 \rightarrow нет;\ \]

\[\frac{1}{13} + 1\frac{3}{13} = 1\frac{4}{13};\ \]

\[1\frac{4}{13} + 2 = 3\frac{4}{13}\text{.\ }\]

\[\textbf{г)}\ a = \frac{3}{13}:\]

\[\frac{3}{13} + \frac{4}{13} = \frac{7}{13};\ \]

\[\frac{7}{13} + \frac{1}{13} = \frac{8}{13};\ \]

\[\frac{8}{13} - \frac{8}{13} = 0 > 1 \rightarrow нет;\ \]

\[0 = 1 \rightarrow нет;\ \]

\[0 + 1\frac{3}{13} = 1\frac{3}{13};\ \]

\[1\frac{3}{13} + 2 = 3\frac{3}{13}\text{.\ }\]

\[\textbf{д)}\ a = 1\frac{5}{13}:\ \]

\[1\frac{5}{13} + \frac{4}{13} = 1\frac{9}{13};\ \]

\[1\frac{9}{13} + \frac{1}{13} = 1\frac{10}{13};\ \]

\[1\frac{10}{13} - \frac{8}{13} = 1\frac{2}{13} > 1 \rightarrow да;\ \]

\[1\frac{2}{13} - 1 = \frac{2}{13};\ \]

\[\frac{2}{13} + 3\frac{2}{13} = 3\frac{4}{13}\text{.\ \ \ }\]