ВАРИАНТ 3. 1. Найдите значение выражения: a) 3 5/8+1 1/3; б) 4 4/9-2 5/6; в) 6 7/12+(5 3/40-4/15).

а) $$3\frac{5}{8}+1\frac{1}{3} = \frac{3 \cdot 8 + 5}{8} + \frac{1 \cdot 3 + 1}{3} = \frac{29}{8} + \frac{4}{3} = \frac{29 \cdot 3}{8 \cdot 3} + \frac{4 \cdot 8}{3 \cdot 8} = \frac{87}{24} + \frac{32}{24} = \frac{87+32}{24} = \frac{119}{24} = 4\frac{23}{24}$$

б) $$4\frac{4}{9}-2\frac{5}{6} = \frac{4 \cdot 9 + 4}{9} - \frac{2 \cdot 6 + 5}{6} = \frac{40}{9} - \frac{17}{6} = \frac{40 \cdot 2}{9 \cdot 2} - \frac{17 \cdot 3}{6 \cdot 3} = \frac{80}{18} - \frac{51}{18} = \frac{80-51}{18} = \frac{29}{18} = 1\frac{11}{18}$$

в) $$6\frac{7}{12}+(5\frac{3}{40}-\frac{4}{15}) = 6\frac{7}{12}+(\frac{5 \cdot 40 + 3}{40}-\frac{4}{15}) = 6\frac{7}{12}+(\frac{203}{40}-\frac{4}{15}) = 6\frac{7}{12}+(\frac{203 \cdot 3}{40 \cdot 3}-\frac{4 \cdot 8}{15 \cdot 8}) = 6\frac{7}{12}+(\frac{609}{120}-\frac{32}{120}) = 6\frac{7}{12}+\frac{609-32}{120} = 6\frac{7}{12}+\frac{577}{120} = 6\frac{7 \cdot 10}{12 \cdot 10}+\frac{577}{120} = 6\frac{70}{120}+\frac{577}{120} = 6 + \frac{70+577}{120} = 6 + \frac{647}{120} = 6 + 5\frac{47}{120} = 11\frac{47}{120}$$

Ответ: а) $$4\frac{23}{24}$$; б) $$1\frac{11}{18}$$; в) $$11\frac{47}{120}$$