Решебник по математике 6 класс Виленкин Часть 1, 2 Часть 1. Параграф 2. Действия со смешанными числами Задание 169

\[\boxed{\mathbf{2.}\mathbf{169}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \frac{5^{\backslash 5}}{6} - \frac{5^{\backslash 3}}{10} = \frac{25}{30} - \frac{15}{30} = \frac{10}{30} =\]

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

\[\textbf{б)}\ \frac{3^{\backslash 7}}{20} - \frac{3^{\backslash 5}}{28} = \frac{21}{140} - \frac{15}{140} =\]

\[= \frac{6}{140} = \frac{3}{70}\]

\[\textbf{в)}\ \frac{3^{\backslash 7}}{4} - \frac{1^{\backslash 2}}{14} = \frac{21}{28} - \frac{2}{28} = \frac{19}{28}\]

\[\textbf{г)}\ \frac{7^{\backslash 13}}{15} - \frac{2^{\backslash 5}}{39} = \frac{91}{195} - \frac{10}{195} =\]

\[= \frac{81}{195} = \frac{27}{65}\]

\[\textbf{д)}\ \frac{26^{\backslash 4}}{33} - \frac{7^{\backslash 3}}{44} = \frac{104}{132} - \frac{21}{132} =\]

\[= \frac{83}{132}\]

\[\textbf{е)}\ \frac{11^{\backslash 2}}{21} - \frac{3^{\backslash 3}}{14} = \frac{22}{42} - \frac{9}{42} = \frac{13}{42}\]

\[\textbf{ж)}\ \frac{9^{\backslash 13}}{22} - \frac{7^{\backslash 11}}{26} = \frac{117}{286} - \frac{77}{286} =\]

\[= \frac{40}{286} = \frac{20}{143}\]

\[\textbf{з)}\ \frac{33^{\backslash 3}}{40} - \frac{7^{\backslash 8}}{15} = \frac{99}{120} - \frac{56}{120} =\]

\[= \frac{43}{120}\]