ГДЗ по математике 6 класс Виленкин Часть 1, 2 Часть 1. Параграф 2. Действия со смешанными числами Задание 386

\[\boxed{\mathbf{2.}\mathbf{386}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \frac{3}{7}a + \frac{2}{7}a = \left( \frac{3}{7} + \frac{2}{7} \right)a = \frac{5}{7}a\]

\[\textbf{б)}\ \frac{9}{14}n - \frac{3}{14}n = \left( \frac{9}{14} - \frac{3}{14} \right)n =\]

\[= \frac{6}{14}n = \frac{3}{7}n\]

\[\textbf{в)}\ \frac{7^{\backslash 2}}{9}c - \frac{11}{18}c = \frac{14}{18}c - \frac{11}{18}c =\]

\[= \left( \frac{14}{18} - \frac{11}{18} \right)c = \frac{3}{18}c = \frac{1}{6}c\]

\[\textbf{г)}\ \frac{7^{\backslash 3}}{8}x - \frac{5^{\backslash 4}}{6}x = \frac{21}{24}x - \frac{20}{24}x =\]

\[= \left( \frac{21}{24} - \frac{20}{24} \right)x = \frac{1}{24}x\]

\[\textbf{д)}\ \frac{4}{13}p + \frac{9}{13}p = \left( \frac{4}{13} + \frac{9}{13} \right)p =\]

\[= \frac{13}{13}p = p\]

\[\textbf{е)}\ \frac{4}{11}a + a = \left( \frac{4}{11} + 1 \right)a =\]

\[= 1\frac{4}{11}a\]

\[\textbf{ж)}\ z - \frac{1}{9}z = \left( 1 - \frac{1}{9} \right)z = \frac{8}{9}z\]

\[\textbf{з)}\ 1\frac{3}{4}t - \frac{7}{8}t = \frac{7^{\backslash 2}}{4}t - \frac{7}{8}t =\]

\[= \frac{14}{8}t - \frac{7}{8}t = \left( \frac{14}{8} - \frac{7}{8} \right)t = \frac{7}{8}t\]