Решение:
a) $$3\frac{3}{4} + 2\frac{2}{3} = \frac{3*4+3}{4} + \frac{2*3+2}{3} = \frac{15}{4} + \frac{8}{3} = \frac{15*3 + 8*4}{12} = \frac{45 + 32}{12} = \frac{77}{12} = 6\frac{5}{12}$$
б) $$11\frac{5}{12} - 5\frac{6}{6} = 11\frac{5}{12} - 6 = 5\frac{5}{12}$$
в) $$4\frac{4}{9} + 1\frac{1}{6} = \frac{4*9+4}{9} + \frac{1*6+1}{6} = \frac{40}{9} + \frac{7}{6} = \frac{40*2 + 7*3}{18} = \frac{80 + 21}{18} = \frac{101}{18} = 5\frac{11}{18}$$
г) $$6\frac{6}{7} - 2\frac{2}{3} = \frac{6*7+6}{7} - \frac{2*3+2}{3} = \frac{48}{7} - \frac{8}{3} = \frac{48*3 - 8*7}{21} = \frac{144 - 56}{21} = \frac{88}{21} = 4\frac{4}{21}$$
д) $$3\frac{3}{10} + 2\frac{2}{15} = \frac{3*10+3}{10} + \frac{2*15+2}{15} = \frac{33}{10} + \frac{32}{15} = \frac{33*3 + 32*2}{30} = \frac{99 + 64}{30} = \frac{163}{30} = 5\frac{13}{30}$$
e) $$7\frac{7}{12} - 3\frac{3}{8} = \frac{7*12+7}{12} - \frac{3*8+3}{8} = \frac{91}{12} - \frac{27}{8} = \frac{91*2 - 27*3}{24} = \frac{182 - 81}{24} = \frac{101}{24} = 4\frac{5}{24}$$
Ответ: a) $$6\frac{5}{12}$$, б) $$5\frac{5}{12}$$, в) $$5\frac{11}{18}$$, г) $$4\frac{4}{21}$$, д) $$5\frac{13}{30}$$, e) $$4\frac{5}{24}$$