ГДЗ по математике 6 класс Мерзляк Задание 450

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Год:2023
Тип:учебник
Серия:Алгоритм успеха
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Задание 450

\[\boxed{\mathbf{450\ (450).}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ 3\frac{3}{4}\ :\frac{3}{8}\ :1\frac{3}{7} = \frac{18}{4}\ :\frac{3}{8}\ :\frac{10}{7} =\]

\[= \frac{18}{4} \cdot \frac{8}{3} \cdot \frac{7}{10} = \frac{18 \cdot 8 \cdot 7}{4 \cdot 3 \cdot 10} = 7\]

\[2)\ 3\ \frac{3}{4}\ :\left( \frac{3}{8} \cdot 1\frac{3}{7} \right) =\]

\[= \frac{15}{4}\ :\left( \frac{3 \cdot 7}{8 \cdot 10} \right) = \frac{15 \cdot 8 \cdot 10}{4 \cdot 3 \cdot 7} =\]

\[= \frac{100}{7} = 14\frac{2}{7}\]

\[3)\ 1\frac{7}{9} \cdot \frac{15}{32}\ :1\frac{19}{36} =\]

\[= \frac{16}{9} \cdot \frac{15}{32}\ :\frac{55}{36} = \frac{16 \cdot 15 \cdot 36}{9 \cdot 32 \cdot 55} =\]

\[= \frac{6}{11}\]

\[4)\ 1\frac{7}{9} \cdot \left( \frac{15}{32}\ :1\frac{19}{36} \right) =\]

\[= \frac{16}{9} \cdot \left( \frac{15}{32}\ :\frac{55}{36} \right) =\]

\[= \frac{16}{9} \cdot \left( \frac{15 \cdot 36}{32 \cdot 55} \right) = \frac{16 \cdot 15 \cdot 36}{9 \cdot 32 \cdot 55} =\]

\[= \frac{6}{11}\]

\[5)\ 3\frac{4}{7}\ :1\frac{1}{7} \cdot \frac{2}{3} = \frac{25}{7}\ :\frac{8}{7} \cdot \frac{2}{3} =\]

\[= \frac{25}{7} \cdot \frac{7}{8} \cdot \frac{2}{3} = \frac{25 \cdot 7 \cdot 2}{7 \cdot 8 \cdot 3} = \frac{25}{12} =\]

\[= 2\frac{1}{12}\]

\[6)\ 3\frac{4}{7}\ :\left( 1\frac{1}{7} \cdot \frac{2}{3} \right) =\]

\[= \frac{25}{7}\ :\left( \frac{8}{7} \cdot \frac{2}{3} \right) = \frac{25}{7} \cdot \left( \frac{7 \cdot 3}{8 \cdot 2} \right) =\]

\[= \frac{25 \cdot 7 \cdot 3}{7 \cdot 8 \cdot 2} = \frac{75}{16} = 4\frac{11}{16}\]

\[7)\ \left( \frac{5}{12} + \frac{1}{8} \right)\ :\frac{3}{8} =\]

\[= \frac{5}{12}\ :\frac{3}{8} + \frac{1}{8}\ :\frac{3}{8} =\]

\[= \frac{5 \cdot 8}{12 \cdot 3} + \frac{1 \cdot 8}{8 \cdot 3} =\]

\[= \frac{10}{9} + \frac{1^{\backslash 3}}{3} = \frac{10 + 3}{9} = \frac{13}{9} = 1\frac{4}{9}\]

\[8)\ \frac{5}{12} + \frac{1}{8}\ :\frac{3}{8} = \frac{5}{12} + \frac{1 \cdot 8}{8 \cdot 3} =\]

\[= \frac{5}{12} + \frac{1^{\backslash 4}}{3} = \frac{5 + 4}{12} = \frac{9}{12} = \frac{3}{4}\]

\[9)\ 2\frac{6}{7}\ :\left( \frac{5^{\backslash 7}}{6} - \frac{9^{\backslash 3}}{14} \right) =\]

\[= \frac{20}{7}\ :\left( \frac{35 - 27}{42} \right) = \frac{20}{7}\ :\frac{8}{42} =\]

\[= \frac{20 \cdot 42}{7 \cdot 8} = 15\]

\[10)\ 2\frac{6}{7}\ :\frac{5}{6} - \frac{9}{14} = \frac{20}{7} \cdot \frac{6}{5} - \frac{9}{14} =\]

\[= \frac{20 \cdot 6}{7 \cdot 5} - \frac{9}{14} = \frac{24^{\backslash 2}}{7} - \frac{9}{14} =\]

\[= \frac{48 - 9}{14} = \frac{39}{14} = 2\frac{11}{14}\]

\[11)\ 2\frac{1}{4}\ :1\frac{4}{11} - \frac{3}{8}\ :\frac{7}{8} =\]

\[= \frac{9}{4}\ :\frac{15}{11} - \frac{3}{8} \cdot \frac{8}{7} =\]

\[= \frac{9 \cdot 11}{4 \cdot 15} - \frac{3 \cdot 8}{8 \cdot 7} = \frac{33^{\backslash 7}}{20} - \frac{3^{\backslash 20}}{7} =\]

\[= \frac{231 - 60}{140} = \frac{171}{140} = 1\frac{31}{140}\]

\[12)\ \left( 3\frac{1}{6} - 5\frac{1}{6}\ :4\frac{2}{15} \right) \cdot \frac{3}{92} =\]

\[= \left( \frac{19}{6} - \frac{31}{6}\ :\frac{62}{15} \right) \cdot \frac{3}{92} =\]

\[= \left( \frac{19}{6} - \frac{31 \cdot 15}{6 \cdot 62} \right) \cdot \frac{3}{92} =\]

\[= \left( \frac{19}{6} - \frac{5}{4} \right) \cdot \frac{3}{92} =\]

\[= \frac{19 \cdot 3}{6 \cdot 92} - \frac{5 \cdot 3}{4 \cdot 92} = \frac{19}{184} - \frac{15}{368} =\]

\[= \frac{38 - 15}{368} = \frac{23}{368} = \frac{1}{16}\]

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