ГДЗ по математике 6 класс Виленкин Часть 1, 2 Часть 1. Параграф 2. Действия со смешанными числами Задание 417

\[\boxed{\mathbf{2.}\mathbf{417}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \frac{4^{\backslash 2}}{7}x + \frac{5}{14}x = \frac{8}{14}x + \frac{5}{14}x =\]

\[= \left( \frac{8}{14} + \frac{5}{14} \right)x = \frac{13}{14}x\]

\[x = 5\frac{1}{4} = \frac{21}{4}:\]

\[\frac{13}{14} \cdot \frac{21}{4} = \frac{13 \cdot 3}{2 \cdot 4} = \frac{39}{8} = 4\frac{7}{8}.\]

\[x = \frac{9}{13}:\]

\[\frac{13}{14} \cdot \frac{9}{13} = \frac{9}{14}.\]

\[\textbf{б)}\frac{5}{16}y + y^{\backslash 16} - \frac{3^{\backslash 2}}{8}y =\]

\[= \frac{5}{16}y + \frac{16}{16}y - \frac{6}{16}y = \frac{15}{16}y\]

\[y = 1\frac{1}{15} = \frac{16}{15}:\]

\[\frac{15}{16} \cdot \frac{16}{15} = 1.\]

\[y = 1\frac{7}{8} = \frac{15}{8}:\]

\[\frac{15}{16} \cdot \frac{15}{8} = \frac{225}{128} = 1\frac{97}{128}.\]

\[\textbf{в)}\ \frac{17}{42}c - \frac{2^{\backslash 6}}{7}c + \frac{7}{18}c =\]

\[= \frac{17}{42}c - \frac{12}{42}c + \frac{7}{18}c =\]

\[= \frac{5^{\backslash 3}}{42}c + \frac{7^{\backslash 7}}{18}c =\]

\[= \frac{15}{126}c + \frac{49}{126}c = \frac{64}{126}c = \frac{32}{63}c\]

\[c = 3\frac{1}{2} = \frac{7}{2}:\]

\[\frac{32}{63} \cdot \frac{7}{2} = \frac{32 \cdot 7}{63 \cdot 2} = \frac{16}{9} = 1\frac{7}{9}.\]

\[c = 2\frac{5}{8} = \frac{21}{8}:\]

\[\frac{32}{63} \cdot \frac{21}{8} = \frac{32 \cdot 21}{63 \cdot 8} = \frac{4}{3} = 1\frac{1}{3}.\]

\[\textbf{г)}\ \frac{3^{\backslash 9}}{4}n + \frac{2^{\backslash 12}}{3}n - \frac{4}{18}n =\]

\[= \frac{27}{36}n + \frac{24}{36}n - \frac{8}{36}n =\]

\[= \left( \frac{27}{36} + \frac{24}{36} - \frac{8}{36} \right)n = \frac{43}{36}n\]

\[n = 1\frac{13}{23} = \frac{36}{23}:\]

\[\frac{43}{36} \cdot \frac{36}{23} = \frac{43}{23} = 1\frac{20}{23}.\]

\[n = \frac{6}{41}:\]

\[\frac{43}{36} \cdot \frac{6}{41} = \frac{43}{6 \cdot 41} = \frac{43}{246}.\]