1) $$6\frac{4}{15}-2\frac{1}{9}+1\frac{3}{10}=$$ 2) $$7\frac{2}{3}-5\frac{5}{36}+3\frac{3}{8}=$$ 3) $$10\frac{9}{22}-3\frac{7}{11}-4\frac{8}{33}=$$ 4) $$12\frac{17}{42}-(3\frac{1}{4}+4\frac{1}{6})=$$ 5) $$15-(6\frac{4}{35}-2\frac{3}{14})=$$

Решение:

1. $$6\frac{4}{15}-2\frac{1}{9}+1\frac{3}{10} = \frac{94}{15} - \frac{19}{9} + \frac{13}{10} = \frac{94 \times 6 - 19 \times 30 + 13 \times 9}{90} = \frac{564 - 570 + 117}{90} = \frac{111}{90} = \frac{37}{30} = 1\frac{7}{30}$$
2. $$7\frac{2}{3}-5\frac{5}{36}+3\frac{3}{8} = \frac{23}{3} - \frac{185}{36} + \frac{27}{8} = \frac{23 \times 24 - 185 \times 3 + 27 \times 13.5}{72}$$ - Здесь есть неточность в расчетах, приводим к общему знаменателю 72. \( \frac{23 \times 24}{72} - \frac{185 \times 2}{72} + \frac{27 \times 9}{72} = \frac{552 - 370 + 243}{72} = \frac{425}{72} = 5\frac{65}{72} \)
3. $$10\frac{9}{22}-3\frac{7}{11}-4\frac{8}{33} = \frac{229}{22} - \frac{40}{11} - \frac{140}{33} = \frac{229 \times 3 - 40 \times 6 - 140 \times 2}{66} = \frac{687 - 240 - 280}{66} = \frac{167}{66} = 2\frac{35}{66}$$
4. $$12\frac{17}{42}-(3\frac{1}{4}+4\frac{1}{6}) = 12\frac{17}{42}-(3\frac{3}{12}+4\frac{2}{12}) = 12\frac{17}{42}-7\frac{5}{12} = \frac{541}{42} - \frac{89}{12} = \frac{541 \times 2 - 89 \times 7}{84} = \frac{1082 - 623}{84} = \frac{459}{84} = \frac{153}{28} = 5\frac{13}{28}$$
5. $$15-(6\frac{4}{35}-2\frac{3}{14}) = 15-(\frac{214}{35}-\frac{31}{14}) = 15-(\frac{214 \times 2 - 31 \times 5}{70}) = 15-(\frac{428-155}{70}) = 15 - \frac{273}{70} = 15 - 3\frac{63}{70} = 15 - 3\frac{9}{10} = 11\frac{1}{10}$$

Ответ: 1) $$1\frac{7}{30}$$, 2) $$5\frac{65}{72}$$, 3) $$2\frac{35}{66}$$, 4) $$5\frac{13}{28}$$, 5) $$11\frac{1}{10}$$.