Question

30 БАЛЛОВ!!!!!Помогите решить пожалуйста,математика,6 класс.

Answer 1

Ответ:

▪︎1

$\frac{52}{45} - (3 \frac{3}{5} - 1 \frac{7}{9} ) \times ( \frac{1}{2} - \frac{1}{3} ) = \frac{52}{45} -2 \frac{27 - 35}{45} \times \frac{3 - 2}{6} = \frac{52}{45} - 1 \frac{37}{45} \times \frac{1}{6} = \frac{52}{45} - \frac{82}{45 \times 6} = 1 \frac{7}{45} - \frac{41}{135} = 1 \frac{21 - 41}{135} = 1 - \frac{20}{135} = 1 - \frac{4}{27} = \frac{23}{27}$

▪︎2

$(3 \frac{8}{60} - 2 \frac{5}{24} ) \div 9 \frac{1}{4} \div 0.1 = 1 \frac{16 - 25}{120} \div \frac{37}{4} \div \frac{1}{10} = \frac{37}{40} \times \frac{4}{37} \times 10 = 1$

▪︎3

$(5.625 \times 1 \frac{1}{3} - 3 \frac{11}{18} \div \frac{5}{9} ) \div 1 \frac{2}{3} = (\frac{45}{8} \times \frac{4}{3} - \frac{65 \times 9}{18 \times 5} ) \div \frac{5}{3} = ( \frac{15}{2} - \frac{13}{2} ) \times \frac{3}{5} = 1 \times \frac{3}{5} = \frac{3}{5}$

▪︎4

$\frac{9}{10} + \frac{4}{9} \div (2 - 1 \frac{11}{21} ) - 1 \frac{2}{3} = \frac{9}{10} + \frac{4}{9} \div \frac{21 - 11}{21} - 1 \frac{2}{3} = \frac{9}{10} + \frac{4 \times 21}{9 \times 10} - 1 \frac{2}{3} = \frac{9}{10} + \frac{2 \times 7}{3 \times 5} - 1 \frac{2}{3} = \frac{27 + 28}{30} - 1 \frac{20}{30} = \frac{55}{30} - 1 \frac{20}{30} = 1\frac{15}{30} - 1 \frac{20}{30} = \frac{5}{30} =\frac{1}{6}$

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