Решебник по математике 6 класс Виленкин Часть 1, 2 Часть 1. Параграф 2. Действия со смешанными числами Задание 197

\[\boxed{\mathbf{2.}\mathbf{197}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \frac{1^{\backslash 7}}{5} + \frac{1^{\backslash 5}}{7} = \frac{7}{35} + \frac{5}{35} = \frac{12}{35}\]

\[\textbf{б)}\ \frac{1^{\backslash 7}}{3} + \frac{2^{\backslash 3}}{7} = \frac{7}{21} + \frac{6}{21} = \frac{13}{21}\]

\[\textbf{в)}\ \frac{3^{\backslash 6}}{5} + \frac{5^{\backslash 5}}{6} = \frac{18}{30} + \frac{25}{30} = \frac{43}{30} =\]

\[= 1\frac{13}{30}\]

\[\textbf{г)}\ \frac{8^{\backslash 5}}{9} - \frac{2^{\backslash 9}}{5} = \frac{40}{45} - \frac{18}{45} = \frac{22}{45}\]

\[\frac{22}{45} + \frac{2^{\backslash 9}}{5} = \frac{22}{45} + \frac{18}{45} = \frac{40}{45} = \frac{8}{9}\]

\[\textbf{д)}\ \frac{5}{12} + \frac{1^{\backslash 2}}{6} = \frac{5}{12} + \frac{2}{12} = \frac{7}{12}\]

\[\textbf{е)}\ \frac{3^{\backslash 3}}{5} - \frac{4}{15} = \frac{9}{15} - \frac{4}{15} = \frac{5}{15} =\]

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

\[\frac{1^{\backslash 5}}{3} + \frac{4}{15} = \frac{5}{15} + \frac{4}{15} = \frac{9}{15} = \frac{3}{5}\]

\[\textbf{ж)}\ \frac{19^{\backslash 5}}{21} - \frac{11^{\backslash 7}}{15} = \frac{95}{105} - \frac{77}{105} =\]

\[= \frac{18}{105} = \frac{6}{35}\]

\[\frac{19^{\backslash 5}}{21} - \frac{6^{\backslash 3}}{35} = \frac{95}{105} - \frac{18}{105} =\]

\[= \frac{77}{105} = \frac{11}{15}\]

\[\textbf{з)}\ \frac{5^{\backslash 3}}{42} + \frac{10^{\backslash 2}}{63} = \frac{15}{126} + \frac{20}{126} =\]

\[= \frac{35}{126} = \frac{5}{18}\]

\[\textbf{и)}\ \frac{11^{\backslash 26}}{21} + \frac{2^{\backslash 21}}{26} =\]

\[= \frac{11 \cdot 26}{546} + \frac{42}{546} = \frac{286}{546} + \frac{42}{546} =\]

\[= \frac{328}{546} = \frac{164}{273}\]

\[к)\ \frac{5^{\backslash 5}}{24} - \frac{7^{\backslash 2}}{60} = \frac{25}{120} - \frac{14}{120} =\]

\[= \frac{11}{120}\]

\[\frac{5^{\backslash 5}}{24} - \frac{11}{120} = \frac{25}{120} - \frac{11}{120} =\]

\[= \frac{14}{120} = \frac{7}{60}\]