Решебник по математике 6 класс Виленкин Часть 1, 2 Часть 1. Параграф 3. Отношения и пропорции Задание 38

\[\boxed{\mathbf{3.38}}\]

\[\textbf{а)}\ \frac{3\frac{1}{12} + 2\frac{1}{6} - \frac{1}{8}}{1,4 + 0,6 \cdot 4,5} = \frac{5\frac{1}{8}}{4,1} =\]

\[= \frac{41}{8}\ :\frac{41}{10} = \frac{41}{8} \cdot \frac{10}{41} = \frac{10}{8} = \frac{5}{4} =\]

\[= 1\frac{1}{4} = 1,25\]

\[3\frac{1^{\backslash 2}}{12} + 2\frac{1^{\backslash 4}}{6} - \frac{1^{\backslash 3}}{8} =\]

\[= 3\frac{2}{24} + 2\frac{4}{24} - \frac{3}{24} = 5\frac{3}{24} = 5\frac{1}{8}\]

\[1,4 + 0,6 \cdot 4,5 = 1,4 + 2,7 = 4,1\]

\[\textbf{б)}\ \frac{10\frac{10}{11}\ :12}{2\frac{21}{22}} \cdot 6\frac{1}{2} =\]

\[= \frac{\frac{120}{11} \cdot \frac{1}{12} \cdot \frac{13}{2}}{\frac{65}{22}} = \frac{65}{11} \cdot \frac{22}{65} = 2\]

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

\[= \frac{8\ :\frac{12}{5}}{\frac{21}{4} \cdot \frac{1}{7}}\ :\frac{\frac{15}{7}\ :\frac{5}{7}}{4 \cdot \frac{9}{8}} =\]

\[= \frac{8 \cdot \frac{5}{12}}{\frac{3}{4}}\ :\frac{\frac{15}{7} \cdot \frac{7}{5}}{\frac{9}{2}} = \frac{\frac{10}{3} \cdot \frac{9}{2}}{\frac{3}{4} \cdot 3} =\]

\[= 15\ :\frac{9}{4} = 15 \cdot \frac{4}{9} = \frac{20}{3} = 6\frac{2}{3}\ \]