Решебник по математике 6 класс Виленкин Часть 1, 2 Часть 1. Параграф 2. Действия со смешанными числами Задание 166

\[\boxed{\mathbf{2.}\mathbf{166}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \frac{1^{\backslash 5}}{2} + \frac{1^{\backslash 2}}{5} = \frac{5}{10} + \frac{2}{10} = \frac{7}{10}\]

\[\textbf{б)}\ \frac{1}{21} + \frac{1^{\backslash 3}}{7} = \frac{1}{21} + \frac{3}{21} = \frac{4}{21}\]

\[\textbf{в)}\ \frac{3^{\backslash 17}}{5} + \frac{2^{\backslash 5}}{17} = \frac{51}{85} + \frac{10}{85} = \frac{61}{85}\]

\[\textbf{г)}\ \frac{1^{\backslash 9}}{7} + \frac{7^{\backslash 7}}{9} = \frac{9}{63} + \frac{49}{63} = \frac{58}{63}\]

\[\textbf{д)}\ \frac{5}{7} + 0 = \frac{5}{7}\]

\[\textbf{е)}\ \frac{2^{\backslash 5}}{3} - \frac{3^{\backslash 3}}{5} = \frac{10}{15} - \frac{6}{15} = \frac{4}{15}\]

\[\textbf{ж)}\ \frac{1^{\backslash 13}}{2} - \frac{1^{\backslash 2}}{13} = \frac{13}{26} - \frac{2}{26} = \frac{11}{26}\]

\[\textbf{з)}\ \frac{3^{\backslash 7}}{5} - \frac{4^{\backslash 5}}{7} = \frac{21}{35} - \frac{20}{35} = \frac{1}{35}\]

\[\textbf{и)}\ \frac{5^{\backslash 9}}{7} - \frac{2^{\backslash 7}}{9} = \frac{45}{63} - \frac{14}{63} = \frac{31}{63}\]

\[к)\ \frac{4}{21} - 0 = \frac{4}{21}\]

\[л)\ \frac{4^{\backslash 5}}{7} + \frac{4^{\backslash 7}}{5} = \frac{20}{35} + \frac{28}{35} = \frac{48}{35} =\]

\[= 1\frac{13}{35}\]

\[м)\ \frac{7^{\backslash 7}}{12} + \frac{13^{\backslash 4}}{21} = \frac{49}{84} + \frac{52}{84} =\]

\[= \frac{101}{84} = 1\frac{17}{84}\]