Ответ:
а) \((3\frac{16}{27}+1\frac{17}{27})-(2\frac{25}{36}+1\frac{7}{36})=5\frac{33}{27}-4\frac{32}{36}=6\frac{1}{9}-4\frac{8}{9}=1\frac{2}{9}\).
б) \((3\frac{25}{48}-1\frac{35}{48})+(4\frac{65}{72}-2\frac{68}{72})=1\frac{38}{48}+1\frac{141}{72}=1\frac{19}{24}+2\frac{23}{24}=4\frac{5}{6}\).
в) \((2\frac{8}{35}-1\frac{13}{35})+(3\frac{20}{49}-2\frac{27}{49})=\frac{30}{35}+\frac{140}{49}=\frac{6}{7}+2\frac{6}{49}=3\frac{48}{49}\).
г) \((4\frac{7}{15}+2\frac{11}{15})-(3\frac{7}{30}+1\frac{29}{30})=7\frac{18}{15}-5\frac{36}{30}=8\frac{1}{5}-6\frac{1}{5}=2\).
Ответ: а) \(1\frac{2}{9}\); б) \(4\frac{5}{6}\); в) \(3\frac{48}{49}\); г) \(2\).
