Решебник по математике 6 класс Виленкин Часть 1, 2 Часть 1. Параграф 3. Отношения и пропорции Задание 58

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Пояснение.

Решение.

\[\textbf{а)}\ 3\frac{1}{2}\ :2\frac{1}{7} = 2\frac{1}{3}\ :t\]

\[3\frac{1}{2}t = 2\frac{1}{3} \cdot 2\frac{1}{7}\]

\[\frac{7}{2}t = \frac{7}{3} \cdot \frac{15}{7}\]

\[t = \frac{7}{3} \cdot \frac{15}{7} \cdot \frac{2}{7}\]

\[t = \frac{10}{7} = 1\frac{3}{7}.\]

\[\textbf{б)}\ 3\frac{1}{3}\ :s = 4\frac{2}{3}\ :1\frac{1}{6}\]

\[4\frac{2}{3}s = 1\frac{1}{6} \cdot 3\frac{1}{3}\]

\[\frac{14}{3}s = \frac{7}{6} \cdot \frac{10}{3}\]

\[s = \frac{7}{6} \cdot \frac{10}{3} \cdot \frac{3}{14} = \frac{5}{3 \cdot 2}\]

\[s = \frac{5}{6}.\]

\[\textbf{в)}\ y\ :\frac{2}{3} = 8\frac{1}{6}\ :2\frac{1}{3}\]

\[2\frac{1}{3}y = 8\frac{1}{6} \cdot \frac{2}{3}\]

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

\[y = \frac{49}{3} \cdot \frac{1}{3} \cdot \frac{3}{7}\]

\[y = \frac{7}{3} = 2\frac{1}{3}.\]

\[\textbf{г)}\ 5\frac{1}{7}\ :\frac{6}{7} = z\ :\frac{12}{17}\]

\[\frac{6}{7}z = \frac{12}{17} \cdot 5\frac{1}{7}\]

\[z = \frac{12}{17} \cdot \frac{36}{7} \cdot \frac{7}{6} = \frac{12 \cdot 6}{17} = \frac{72}{17}\]

\[z = 4\frac{4}{17}\text{.\ }\]