ГДЗ по математике 6 класс Ерина Никольский рабочая тетрадь Часть 1, 2 Часть 1 | Страница 90

Авторы:
Тип:рабочая тетрадь
Часть:1, 2
Серия:Учебно-методический комплект

Страница 90

\[\mathbf{Страница\ 90.\ }\]

\[\boxed{\mathbf{4.}}\]

\[\textbf{а)}\ \frac{1}{10}\ :\frac{1}{5} = \frac{1}{10} \cdot 5 = \frac{1}{2}\]

\[3\ :\frac{3}{8} = 3 \cdot \frac{8}{3} = 8\]

\[\frac{4}{7} \cdot 7 = 4\]

\[\frac{3}{11}\ :\frac{11}{7} = \frac{3}{11} \cdot \frac{7}{11} = \frac{21}{121}\]

\[\textbf{б)}\ 8 \cdot \frac{3}{16} = \frac{3}{2} = 1,5\]

\[\frac{2}{15}\ :\frac{1}{7} = \frac{2}{15} \cdot 7 = \frac{14}{15}\]

\[\frac{3}{5} \cdot \frac{1}{27} = \frac{1}{45}\]

\[\frac{1}{12} \cdot \frac{5}{9} = \frac{5}{108}\]

\[\textbf{в)}\ \frac{11}{20} \cdot 2 = \frac{11}{10} = 1,1\]

\[\frac{12}{13}\ :24 = \frac{12}{13} \cdot \frac{1}{24} = \frac{1}{26}\]

\[16 \cdot \frac{3}{4} = 4 \cdot 3 = 12\]

\[0\ :\frac{5}{6} = 0\]

\[\boxed{\mathbf{5.}}\]

\[\textbf{а)}\frac{- 3 \cdot 18}{- 16 \cdot ( - 6) \cdot ( - 5)} =\]

\[= \frac{3 \cdot 3}{16 \cdot 5} = \frac{9}{80}\]

\[\textbf{б)}\ \frac{- 5 \cdot ( - 2) \cdot ( - 3)}{- 2 \cdot ( - 15) \cdot ( - 6)} = \frac{1}{6}\]

\[\textbf{в)}\ \frac{- 56 \cdot ( - 75)}{- 25 \cdot ( - 112)} = \frac{3}{2} = 1,5\]

\[\textbf{г)}\ \frac{- 125 \cdot ( - 96)}{( - 192) \cdot ( - 75)} = \frac{5}{2 \cdot 3} = \frac{5}{6}\]

\[\textbf{д)}\ \frac{228 \cdot ( - 68)}{38 \cdot ( - 272)} = \frac{6}{4} = 1,5\]

\[\textbf{е)}\ \frac{152 \cdot ( - 171)}{228 \cdot 204} = - \frac{2 \cdot 57}{3 \cdot 68} =\]

\[= - \frac{19}{34}\]

\[\textbf{ж)}\ \frac{- 156 \cdot 91}{- 182 \cdot ( - 26)} = - \frac{6}{2} = - 3\]

\[\textbf{з)}\ \frac{- 35 \cdot 132}{90 \cdot ( - 45)} = \frac{7 \cdot 22}{15 \cdot 9} =\]

\[= \frac{154}{135} = 1\frac{19}{135}\]

\[\boxed{\mathbf{6.}}\]

\[\textbf{а)} - \frac{7}{9} \cdot \frac{5}{14} = - \frac{5}{18}\]

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

\[- \frac{3}{10} \cdot \left( - \frac{5}{7} \right) = \frac{3}{14}\ \]

\[- \frac{9}{2} \cdot \left( - \frac{8}{45} \right) = \frac{4}{5}\]

\[\textbf{б)}\ - \frac{3}{11} \cdot \frac{22}{9} = - \frac{2}{3}\]

\[- \frac{5}{14} \cdot \frac{2}{3} = - \frac{5}{21}\]

\[- \frac{7}{13} \cdot \left( - \frac{26}{49} \right) = \frac{2}{7}\]

\[- \frac{8}{7} \cdot \left( - \frac{21}{72} \right) = \frac{3}{9} = \frac{1}{3}\]

\[\boxed{\mathbf{7.}}\]

\[\textbf{а)} - \frac{1}{5} \cdot 3 = - \frac{3}{5}\]

\[8 \cdot \left( - \frac{1}{7} \right) = - \frac{8}{7} = - 1\frac{1}{7}\]

\[- 6 \cdot \left( - \frac{1}{4} \right) = \frac{6}{4} = 1,5\]

\[\textbf{б)} - \frac{1}{7} \cdot 5 = - \frac{5}{7}\]

\[9 \cdot \left( - \frac{1}{8} \right) = - \frac{9}{8} = - 1\frac{1}{8}\]

\[- 8 \cdot \left( - \frac{1}{6} \right) = \frac{8}{6} = \frac{4}{3} = 1\frac{1}{3}\]

\[\boxed{\mathbf{8.}}\]

\[\textbf{а)} - \frac{4}{7}\ :\left( - \frac{7}{8} \right) = \frac{4}{7} \cdot \frac{8}{7} = \frac{32}{49}\]

\[- \frac{18}{24}\ :\left( - \frac{9}{8} \right) = \frac{18}{24} \cdot \frac{8}{9} = \frac{2}{3}\]

\[- \frac{7}{12}\ :\frac{1}{3} = - \frac{7}{12} \cdot 3 = - \frac{7}{4} =\]

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

\[\textbf{б)} - \frac{2}{5}\ :\left( - \frac{5}{6} \right) = \frac{2}{5} \cdot \frac{6}{5} = \frac{12}{25}\]

\[- \frac{12}{27}\ :\left( - \frac{8}{9} \right) = \frac{12}{27} \cdot \frac{9}{8} = \frac{3}{6} = \frac{1}{2}\]

\[- \frac{11}{18}\ :\frac{1}{9} = - \frac{11}{18} \cdot 9 = - \frac{11}{2} =\]

\[= - 5,5\]

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