ГДЗ по математике 6 класс Виленкин Часть 1, 2 Часть 1. Параграф 2. Действия со смешанными числами Задание 468

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Год:2020-2021-2022
Тип:учебник
Часть:1, 2
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Задание 468

\[\boxed{\mathbf{2.468}}\]

\[\textbf{а)}\ \frac{1}{9}x + \frac{4}{9}x = 3\frac{1}{18}\]

\[\frac{5}{9}x = \frac{55}{18}\]

\[x = \frac{55}{18}\ :\frac{5}{9} = \frac{55}{18} \cdot \frac{9}{5}\]

\[x = \frac{11}{2} = 5,5\]

\[Ответ:x = 5,5.\]

\[\textbf{б)}\ \frac{5^{\backslash 3}}{7}y + \frac{2^{\backslash 7}}{3}y - 4 = \frac{1}{7}\]

\[\frac{15}{21}y + \frac{14}{21}y = \frac{1}{7} + 4\]

\[\frac{29}{21}y = 4\frac{1}{7}\]

\[\frac{29}{21}y = \frac{29}{7}\]

\[y = \frac{29}{7}\ :\frac{29}{21} = \frac{29}{7} \cdot \frac{21}{29}\]

\[y = 3\]

\[Ответ:y = 3.\]

\[\textbf{в)}\ n + \frac{5}{14}n = \frac{1}{7}\]

\[1\frac{5}{14}n = \frac{1}{7}\]

\[\frac{19}{14}n = \frac{1}{7}\]

\[n = \frac{1}{7}\ :\frac{19}{14} = \frac{1}{7} \cdot \frac{14}{19}\]

\[n = \frac{2}{19}\]

\[Ответ:n = \frac{2}{19}.\]

\[\textbf{г)}\ y - \frac{1}{9}y = 5\frac{1}{3}\]

\[\frac{8}{9}y = \frac{16}{3}\]

\[y = \frac{16}{3}\ :\frac{8}{9} = \frac{16}{3} \cdot \frac{9}{8} = 2 \cdot 3\]

\[y = 6\]

\[Ответ:y = 6.\]

\[\textbf{д)}\ \frac{2^{\backslash 3}}{7}c + \frac{2^{\backslash 7}}{3}c - \frac{11}{21}c = 3\frac{1}{2}\]

\[\frac{6}{21}c + \frac{14}{21}c - \frac{11}{21}c = \frac{7}{2}\]

\[\frac{9}{21}c = \frac{7}{2}\]

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

\[c = \frac{7}{2}\ :\frac{3}{7} = \frac{7}{2} \cdot \frac{7}{3} = \frac{49}{6}\]

\[c = 8\frac{1}{6}\]

\[Ответ:c = 8\frac{1}{6}.\]

\[\textbf{е)}\ \frac{5}{8}x + x - \frac{3^{\backslash 2}}{4}x = 1\frac{3}{4}\]

\[\frac{5}{8}x + \frac{8}{8}x - \frac{6}{8}x = \frac{7}{4}\]

\[\frac{7}{8}x = \frac{7}{4}\ \]

\[x = \frac{7}{4}\ :\frac{7}{8} = \frac{7}{4} \cdot \frac{8}{7}\]

\[x = 2\]

\[Ответ:x = 2.\]

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