Решебник по математике 6 класс Виленкин ФГОС Часть 1, 2 Часть 1. П. 3. Отношения и пропорции Задание 161

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Пояснение.

Решение.

\[\textbf{а)}\ 3\frac{2}{3}\ :a = 4\frac{8}{9}\ :1\frac{5}{7}\]

\[4\frac{8}{9}a = 3\frac{2}{3} \cdot 1\frac{5}{7}\]

\[\frac{44}{9}a = \frac{11}{3} \cdot \frac{12}{7}\]

\[a = \frac{44}{7} \cdot \frac{9}{44}\]

\[a = \frac{9}{7} = 1\frac{2}{7}.\]

\[\textbf{б)}\ 1\frac{7}{8}\ :2\frac{1}{3} = 3\frac{3}{4}\ :b\]

\[1\frac{7}{8}b = 2\frac{1}{3} \cdot 3\frac{3}{4}\]

\[\frac{15}{8}b = \frac{7}{3} \cdot \frac{15}{4}\]

\[b = \frac{7}{3} \cdot \frac{15}{4} \cdot \frac{8}{15} = \frac{7 \cdot 2}{3} = \frac{14}{3}\]

\[b = 4\frac{2}{3}.\]

\[\textbf{в)}\ 8\frac{1}{4}\ :c = 13\frac{3}{4}\ :2\frac{1}{3}\]

\[13\frac{3}{4}c = 8\frac{1}{4} \cdot 2\frac{1}{3}\]

\[\frac{55}{4}c = \frac{33}{4} \cdot \frac{7}{3}\]

\[c = \frac{77}{4} \cdot \frac{4}{55} = \frac{7}{5}\]

\[c = 1,4.\]

\[\textbf{г)}\ 5\frac{2}{3}\ :2\frac{5}{6} = 2\frac{1}{7}\ :d\]

\[5\frac{2}{3}d = 2\frac{5}{6} \cdot 2\frac{1}{7}\]

\[\frac{17}{3}d = \frac{17}{6} \cdot \frac{15}{7}\ \]

\[d = \frac{17}{2} \cdot \frac{5}{7} \cdot \frac{3}{17} = \frac{15}{14}\]

\[d = 1\frac{1}{14}.\]