5.7 Раскройте скобки и найдите значение суммы: a) \(\frac{5}{9}+(\frac{4}{9}-\frac{7}{11});\) б) \(5\frac{3}{7}+(-\frac{3}{7}-\frac{8}{9});\) в) \(4,32+(\frac{12}{13}-3,32);\) г) \(\frac{7}{15}-(\frac{4}{15}-\frac{4}{5});\) д) \(6\frac{7}{9}-(3\frac{4}{9}+2\frac{1}{3});\) e) \(-9\frac{11}{12}-(\frac{1}{4}-\frac{5}{12});\) ж) \((4\frac{1}{4}-6\frac{8}{11})+(6,75-3\frac{3}{11});\) з) \((9\frac{7}{18}-2,7)-(4\frac{1}{18}+2,3).\)

Ответ:

a) \(\frac{5}{9}+(\frac{4}{9}-\frac{7}{11}) = \frac{5}{9} + \frac{4}{9} - \frac{7}{11} = \frac{55 + 44 - 63}{99} = \frac{99 - 63}{99} = \frac{36}{99} = \frac{4}{11}\)

б) \(5\frac{3}{7}+(-\frac{3}{7}-\frac{8}{9}) = 5\frac{3}{7} - \frac{3}{7} - \frac{8}{9} = 5 - \frac{8}{9} = 4\frac{1}{9}\)

в) \(4,32+(\frac{12}{13}-3,32) = 4,32 + \frac{12}{13} - 3,32 = 1 + \frac{12}{13} = \frac{13+12}{13} = \frac{25}{13} = 1\frac{12}{13}\)

г) \(\frac{7}{15}-(\frac{4}{15}-\frac{4}{5}) = \frac{7}{15} - \frac{4}{15} + \frac{4}{5} = \frac{7-4+12}{15} = \frac{15}{15} = 1\)

д) \(6\frac{7}{9}-(3\frac{4}{9}+2\frac{1}{3}) = 6\frac{7}{9} - 3\frac{4}{9} - 2\frac{1}{3} = 6\frac{7}{9} - 3\frac{4}{9} - 2\frac{3}{9} = 6\frac{7}{9} - 5\frac{7}{9} = 1\)

e) \(-9\frac{11}{12}-(\frac{1}{4}-\frac{5}{12}) = -9\frac{11}{12} - \frac{1}{4} + \frac{5}{12} = -9\frac{11}{12} - \frac{3}{12} + \frac{5}{12} = -9\frac{11 - 3 + 5}{12} = -9\frac{13}{12} = -10\frac{1}{12}\)

ж) \((4\frac{1}{4}-6\frac{8}{11})+(6,75-3\frac{3}{11}) = 4\frac{1}{4} - 6\frac{8}{11} + 6,75 - 3\frac{3}{11} = 4,25 - 6\frac{8}{11} + 6,75 - 3\frac{3}{11} = 4,25 + 6,75 - 6\frac{8}{11} - 3\frac{3}{11} = 11 - 9\frac{11}{11} = 11 - 10 = 1\)

з) \((9\frac{7}{18}-2,7)-(4\frac{1}{18}+2,3) = 9\frac{7}{18} - 2,7 - 4\frac{1}{18} - 2,3 = 9\frac{7}{18} - 4\frac{1}{18} - 2,7 - 2,3 = 5\frac{6}{18} - 5 = \frac{6}{18} = \frac{1}{3}\)

Ответ: \(\frac{4}{11}; 4\frac{1}{9}; 1\frac{12}{13}; 1; 1; -10\frac{1}{12}; 1; \frac{1}{3}\)

Ответ: