2. Упростите выражение: a) $$2\frac{5}{7}a + \frac{14}{5}a - \frac{20}{21}a;$$ б) $$3\frac{2}{3}n - 2\frac{4}{9}n + 1\frac{1}{2}n$$ в) $$\frac{3}{8}k + \frac{4}{9}k - \frac{5}{12}k;$$ г) $$1\frac{5}{12}x + 2\frac{7}{18}x - 1\frac{1}{2}x$$

Упростим каждое выражение.

1. a) $$2\frac{5}{7}a + \frac{14}{5}a - \frac{20}{21}a = \frac{19}{7}a + \frac{14}{5}a - \frac{20}{21}a = a(\frac{19}{7} + \frac{14}{5} - \frac{20}{21}) = a(\frac{19 \cdot 15}{7 \cdot 15} + \frac{14 \cdot 21}{5 \cdot 21} - \frac{20 \cdot 5}{21 \cdot 5}) = a(\frac{285}{105} + \frac{294}{105} - \frac{100}{105}) = a\frac{285+294-100}{105} = a\frac{479}{105} = 4\frac{59}{105}a$$

   Перевод: $$2\frac{5}{7}a + \frac{14}{5}a - \frac{20}{21}a = \frac{19}{7}a + \frac{14}{5}a - \frac{20}{21}a = a(\frac{19}{7} + \frac{14}{5} - \frac{20}{21}) = a(\frac{19 \cdot 15}{7 \cdot 15} + \frac{14 \cdot 21}{5 \cdot 21} - \frac{20 \cdot 5}{21 \cdot 5}) = a(\frac{285}{105} + \frac{294}{105} - \frac{100}{105}) = a\frac{285+294-100}{105} = a\frac{479}{105} = 4\frac{59}{105}a$$

   Ответ: $$4\frac{59}{105}a$$

2. б) $$3\frac{2}{3}n - 2\frac{4}{9}n + 1\frac{1}{2}n = \frac{11}{3}n - \frac{22}{9}n + \frac{3}{2}n = n(\frac{11}{3} - \frac{22}{9} + \frac{3}{2}) = n(\frac{11 \cdot 6}{3 \cdot 6} - \frac{22 \cdot 2}{9 \cdot 2} + \frac{3 \cdot 9}{2 \cdot 9}) = n(\frac{66}{18} - \frac{44}{18} + \frac{27}{18}) = n\frac{66-44+27}{18} = n\frac{49}{18} = 2\frac{13}{18}n$$

   Перевод: $$3\frac{2}{3}n - 2\frac{4}{9}n + 1\frac{1}{2}n = \frac{11}{3}n - \frac{22}{9}n + \frac{3}{2}n = n(\frac{11}{3} - \frac{22}{9} + \frac{3}{2}) = n(\frac{11 \cdot 6}{3 \cdot 6} - \frac{22 \cdot 2}{9 \cdot 2} + \frac{3 \cdot 9}{2 \cdot 9}) = n(\frac{66}{18} - \frac{44}{18} + \frac{27}{18}) = n\frac{66-44+27}{18} = n\frac{49}{18} = 2\frac{13}{18}n$$

   Ответ: $$2\frac{13}{18}n$$

3. в) $$\frac{3}{8}k + \frac{4}{9}k - \frac{5}{12}k = k(\frac{3}{8} + \frac{4}{9} - \frac{5}{12}) = k(\frac{3 \cdot 9}{8 \cdot 9} + \frac{4 \cdot 8}{9 \cdot 8} - \frac{5 \cdot 6}{12 \cdot 6}) = k(\frac{27}{72} + \frac{32}{72} - \frac{30}{72}) = k\frac{27+32-30}{72} = k\frac{29}{72}$$

   Перевод: $$\frac{3}{8}k + \frac{4}{9}k - \frac{5}{12}k = k(\frac{3}{8} + \frac{4}{9} - \frac{5}{12}) = k(\frac{3 \cdot 9}{8 \cdot 9} + \frac{4 \cdot 8}{9 \cdot 8} - \frac{5 \cdot 6}{12 \cdot 6}) = k(\frac{27}{72} + \frac{32}{72} - \frac{30}{72}) = k\frac{27+32-30}{72} = k\frac{29}{72}$$

   Ответ: $$\frac{29}{72}k$$

4. г) $$1\frac{5}{12}x + 2\frac{7}{18}x - 1\frac{1}{2}x = \frac{17}{12}x + \frac{43}{18}x - \frac{3}{2}x = x(\frac{17}{12} + \frac{43}{18} - \frac{3}{2}) = x(\frac{17 \cdot 3}{12 \cdot 3} + \frac{43 \cdot 2}{18 \cdot 2} - \frac{3 \cdot 18}{2 \cdot 18}) = x(\frac{51}{36} + \frac{86}{36} - \frac{54}{36}) = x\frac{51+86-54}{36} = x\frac{83}{36} = 2\frac{11}{36}x$$

   Перевод: $$1\frac{5}{12}x + 2\frac{7}{18}x - 1\frac{1}{2}x = \frac{17}{12}x + \frac{43}{18}x - \frac{3}{2}x = x(\frac{17}{12} + \frac{43}{18} - \frac{3}{2}) = x(\frac{17 \cdot 3}{12 \cdot 3} + \frac{43 \cdot 2}{18 \cdot 2} - \frac{3 \cdot 18}{2 \cdot 18}) = x(\frac{51}{36} + \frac{86}{36} - \frac{54}{36}) = x\frac{51+86-54}{36} = x\frac{83}{36} = 2\frac{11}{36}x$$

   Ответ: $$2\frac{11}{36}x$$