Решебник по математике 6 класс Мерзляк ФГОС Задание 1347

 

\[\boxed{\mathbf{1347}\mathbf{.}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[3)\ \left( 5\frac{5}{28} - 5\frac{1}{3} \cdot 1,25 - 1\frac{16}{21} \right)\ :\]

\[:( - 1,5) = 2\frac{1}{6}\]

\[5\frac{1}{3} \cdot 1\frac{25}{100} = 5\frac{1}{3} \cdot 1\frac{1}{4} = \frac{16 \cdot 5}{3 \cdot 4} =\]

\[= \frac{20}{3} = 6\frac{2}{3}\]

\[5\frac{5}{28} - 6\frac{2}{3} - 1\frac{16}{21} = 5\frac{15}{84} -\]

\[- 6\frac{56}{84} - 1\frac{64}{84} = - 1\frac{41}{84} -\]

\[- 1\frac{64}{84} = - 2\frac{105}{84} =\]

\[= - 3\frac{21}{84} = - 3\frac{1}{4}\]

\[- 3\frac{1}{4}\ :( - 1,5) = \frac{13}{4}\ :\frac{3}{2} =\]

\[= \frac{13 \cdot 2}{4 \cdot 3} = \frac{13}{6} = 2\frac{1}{6}\]

\[4)\ \left( - 3\frac{1}{3} \cdot 1,9 + 19,5\ :4\frac{1}{3} \right)\ :\]

\[:\left( 0,16 - \frac{62}{75} \right) = 2\frac{3}{4}\]

\[- 3\frac{1}{3} \cdot 1\frac{9}{10} = - \frac{10}{3} \cdot \frac{19}{10} =\]

\[= - \frac{19}{3} = - 6\frac{1}{3}\]

\[19\frac{5}{10}\ :4\frac{1}{3} = \frac{195}{10}\ :\frac{13}{3} =\]

\[= \frac{195 \cdot 3}{10 \cdot 13} = \frac{45}{10} = 4\frac{1}{2}\]

\[- 6\frac{1}{3} + 4\frac{1}{2} = - 6\frac{2}{6} + 4\frac{3}{6} =\]

\[= - 5\frac{8}{6} + 4\frac{3}{6} = - 1\frac{5}{6}\]

\[0,16 - \frac{62}{75} = \frac{16}{100} - \frac{62}{75} = \frac{4}{25} -\]

\[- \frac{62}{75} = \frac{12}{75} - \frac{62}{75} = - \frac{50}{75} = - \frac{2}{3}\]

\[- 1\frac{5}{6}\ :\left( - \frac{2}{3} \right) = \frac{11}{6}\ :\frac{2}{3} = \frac{11 \cdot 3}{6 \cdot 2} =\]

\[= \frac{11}{4} = 2\frac{3}{4}\]

\[5)\ \frac{- 2\frac{2}{11} \cdot 4,125 + 1,6 \cdot 3\frac{3}{4}}{9 - 5\frac{5}{6} \cdot 2\frac{4}{7}} = 0,5\]

\[- 2\frac{2}{11} \cdot 4\frac{1}{8} = - \frac{24}{11} \cdot \frac{33}{8} =\]

\[= - 3 \cdot 3 = - 9\]

\[1,6 \cdot 3\frac{3}{4} = 1\frac{3}{5} \cdot 3\frac{3}{4} =\]

\[= \frac{8 \cdot 15}{5 \cdot 4} = 2 \cdot 3 = 6\]

\[- 9 + 6 = - 3\]

\[5\frac{5}{6} \cdot 2\frac{4}{7} = \frac{35 \cdot 18}{6 \cdot 7} = 5 \cdot 3 = 15\]

\[9 - 15 = - 6\]

\[\frac{- 3}{- 6} = \frac{1}{2} = 0,5\]

\[6)\ \frac{- 2\frac{7}{24}\ :1\frac{5}{6} - 1,6 \cdot ( - 0,3)}{- 9,5\ :\left( 5\frac{7}{10} - 4\frac{12}{35} \right)} =\]

\[= 0,11\]

\[- 2\frac{7}{24}\ :1\frac{5}{6} = - \frac{55}{24}\ :\frac{11}{6} =\]

\[= - \frac{55 \cdot 6}{24 \cdot 11} = - \frac{5}{4} = - 1,25\]

\[1,6 \cdot ( - 0,3) = - 0,48\]

\[- 1,25 - ( - 0,48) = - 1,25 +\]

\[+ 0,48 = - 0,77\]

\[5\frac{7}{10} - 4\frac{12}{35} = 5\frac{49}{70} - 4\frac{24}{70} =\]

\[= 1\frac{25}{70} = 1\frac{5}{14}\]

\[- 9,5\ :1\frac{5}{14} = - \frac{95}{10}\ :\frac{19}{14} =\]

\[= - \frac{95 \cdot 14}{10 \cdot 19} = - 7\]

\[- 0,77\ :( - 7) = 0,11\]

\[7)\ \frac{- 0,4 \cdot \left( - 6,3\ :3,15 + \frac{5}{6} \cdot 0,9 \right)}{- 48 - \frac{2}{7} \cdot ( - 91)} = - \frac{1}{44}\]

\[- 6,3\ :3,15 = - 630\ :315 = - 2\]

\[\frac{5}{6} \cdot 0,9 = \frac{5}{6} \cdot \frac{9}{10} = \frac{3}{4}\]

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

\[- 0,4 \cdot \left( - 1\frac{1}{4} \right) = \frac{2}{5} \cdot \frac{5}{4} = \frac{1}{2}\]

\[\frac{2}{7} \cdot ( - 91) = 2 \cdot ( - 13) = - 26\]

\[- 48 - ( - 26) =\]

\[= - 48 + 26 = - 22\]

\[\frac{1}{2}\ :( - 22) = - \frac{1}{2} \cdot \frac{1}{22} = - \frac{1}{44}\]

\[8)\ ( - 13,6 + 5,1) \cdot 1\frac{3}{17} +\]

\[+ \left( 2\frac{7}{23} - 1\frac{45}{46} \right)\ :1\frac{7}{23} = - 9,75\]

\[- 13,6 + 5,1 = - 8,5\]

\[- 8,5 \cdot 1\frac{3}{17} = - \frac{85}{10} \cdot \frac{20}{17} =\]

\[= - 5 \cdot 2 = 10\]

\[2\frac{7}{23} - 1\frac{45}{46} = 2\frac{14}{46} - 1\frac{45}{46} =\]

\[= 1\frac{60}{46} - 1\frac{45}{46} = \frac{15}{46}\]

\[\frac{15}{46}\ :1\frac{7}{23} = \frac{15}{46}\ :\frac{30}{23} =\]

\[= \frac{15}{46} \cdot \frac{23}{30} = \frac{1}{4}\]

\[- 10 + \frac{1}{4} = - 9\frac{3}{4} = - 9,75\]