Решебник по математике 6 класс Виленкин Часть 1, 2 Часть 1. Параграф 3. Отношения и пропорции Задание 42

\[\boxed{\mathbf{3.42}}\]

Пояснение.

Решение.

\[\textbf{а)}\ t\ :42,4 = 26,1\ :63,6\]

\[63,6t = 42,4 \cdot 26,1\]

\[t = \frac{424 \cdot 261}{636} = \frac{106 \cdot 261}{159} =\]

\[= \frac{106 \cdot 87}{53} = 2 \cdot 87\]

\[t = 174.\]

\[\textbf{б)}\ 4\frac{1}{2}\ :2\frac{2}{5} = 3\frac{1}{4}\ :t\]

\[4\frac{1}{2}t = 3\frac{1}{4} \cdot 2\frac{2}{5}\]

\[t = \frac{13}{4} \cdot \frac{12}{5}\ :\frac{9}{2} = \frac{13 \cdot 12 \cdot 2}{4 \cdot 5 \cdot 9} =\]

\[= \frac{13 \cdot 2}{3} = \frac{26}{3}\]

\[t = 8\frac{2}{3}.\]

\[\textbf{в)}\ 4,5\ :2,5 = y\ :3,5\]

\[2,5y = 3,5 \cdot 4,5\]

\[y = \frac{35 \cdot 45}{25} = 7 \cdot 9\]

\[y = 63.\]

\[\textbf{г)}\ \frac{25}{6}\ :x = \frac{20}{21}\ :\frac{4}{7}\]

\[\frac{20}{21}x = \frac{4}{7} \cdot \frac{25}{6}\]

\[x = \frac{4}{7} \cdot \frac{25}{6} \cdot \frac{21}{20} = \frac{4 \cdot 25 \cdot 21}{7 \cdot 6 \cdot 20} =\]

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

\[x = 2,5.\]