а) $$1\frac{7}{12}+3\frac{9}{16}$$; НОК (12, 16) = 48 $$1\frac{7}{12}+3\frac{9}{16}=1+\frac{7}{12}+3+\frac{9}{16}=(1+3)+(\frac{7}{12}+\frac{9}{16})=$$ $$=\frac{}{48}+\frac{}{48}=4+\frac{}{48}=4+\frac{}{48}+1=\frac{}{48}=5\frac{}{48}$$ б) $$2\frac{3}{4}+4\frac{5}{14}$$; НОК (4, 14) = $$2\frac{3}{4}+4\frac{5}{14}=2+\frac{}{}+4+\frac{}{}=(2 + 4) + (\frac{}{}+\frac{}{})=$$ 2. Вычислите разность: а) $$3\frac{7}{10}-1\frac{4}{15}$$; НОК (10, 15) = $$3\frac{7}{10}-1\frac{4}{15}=3-\frac{}{}-1-\frac{}{}= (3 - 1) + (\frac{}{}-\frac{}{})=2+\frac{}{}=$$ $$2+=$$ б) $$5\frac{1}{12}-2\frac{7}{15}$$; НОК (12, 15) = 60, $$5\frac{1}{12}-2\frac{7}{15}=5\frac{()}{}-2\frac{()}{}=\frac{5}{60}-\frac{28}{60}=(4+1+\frac{5}{60})-2\frac{28}{60}=$$ $$=(4+\frac{60}{60}+\frac{5}{60})-2\frac{28}{60}=4\frac{65}{60}-2\frac{28}{60}=2\frac{65-28}{60}=2\frac{}{60}=$$ в) $$6\frac{9}{35}-5\frac{13}{20}$$; НОК (35, 20) = $$6\frac{9}{35}-5\frac{13}{20}=$$ г) $$4\frac{7}{30}-3\frac{5}{42}$$; НОК (30, 42) = $$4\frac{7}{30}-3\frac{5}{42}=$$ 3. Найдите разность, представив смешанные числа в виде неправильных дробей: a) $$4\frac{5}{6}-2\frac{3}{8}=\frac{29}{6}-\frac{}{8}=\frac{}{24}-\frac{}{24}=$$ б) $$5\frac{1}{9}-3\frac{5}{12}=$$\n

1. a) $$1\frac{7}{12}+3\frac{9}{16}$$; НОК (12, 16) = 48 $$1\frac{7}{12}+3\frac{9}{16}=1+\frac{7}{12}+3+\frac{9}{16}=(1+3)+(\frac{7}{12}+\frac{9}{16}) = 4 + (\frac{7\cdot4}{12\cdot4} + \frac{9\cdot3}{16\cdot3}) =$$ $$=4+\frac{28}{48}+\frac{27}{48}=4+\frac{55}{48}=4+\frac{48}{48}+\frac{7}{48}=5\frac{7}{48}$$ б) $$2\frac{3}{4}+4\frac{5}{14}$$; НОК (4, 14) = 28 $$2\frac{3}{4}+4\frac{5}{14}=2+\frac{3}{4}+4+\frac{5}{14}=(2+4)+(\frac{3}{4}+\frac{5}{14}) = 6 + (\frac{3\cdot7}{4\cdot7} + \frac{5\cdot2}{14\cdot2}) = 6 + (\frac{21}{28} + \frac{10}{28}) = 6 + \frac{31}{28} = 6 + \frac{28}{28} + \frac{3}{28} = 7\frac{3}{28}$$ 2. a) $$3\frac{7}{10}-1\frac{4}{15}$$; НОК (10, 15) = 30 $$3\frac{7}{10}-1\frac{4}{15}=3+\frac{7}{10}-1-\frac{4}{15} = (3 - 1) + (\frac{7}{10}-\frac{4}{15}) = 2 + (\frac{7\cdot3}{10\cdot3} - \frac{4\cdot2}{15\cdot2}) = 2 + (\frac{21}{30} - \frac{8}{30}) = 2 + \frac{13}{30} = 2\frac{13}{30}$$ б) $$5\frac{1}{12}-2\frac{7}{15}$$; НОК (12, 15) = 60, $$5\frac{1}{12}-2\frac{7}{15}=5\frac{1\cdot5}{12\cdot5}-2\frac{7\cdot4}{15\cdot4}=\frac{5}{60}-\frac{28}{60}=(4+1+\frac{5}{60})-2\frac{28}{60}=$$ $$=(4+\frac{60}{60}+\frac{5}{60})-2\frac{28}{60}=4\frac{65}{60}-2\frac{28}{60}=2\frac{65-28}{60}=2\frac{37}{60}$$ в) $$6\frac{9}{35}-5\frac{13}{20}$$; НОК (35, 20) = 140 $$6\frac{9}{35}-5\frac{13}{20} = 6 + \frac{9}{35} - 5 - \frac{13}{20} = (6 - 5) + (\frac{9}{35} - \frac{13}{20}) = 1 + (\frac{9\cdot4}{35\cdot4} - \frac{13\cdot7}{20\cdot7}) = 1 + (\frac{36}{140} - \frac{91}{140}) = 1 - \frac{55}{140} = \frac{140}{140} - \frac{55}{140} = \frac{85}{140} = \frac{17}{28}$$ г) $$4\frac{7}{30}-3\frac{5}{42}$$; НОК (30, 42) = 210 $$4\frac{7}{30}-3\frac{5}{42} = 4 + \frac{7}{30} - 3 - \frac{5}{42} = (4 - 3) + (\frac{7}{30} - \frac{5}{42}) = 1 + (\frac{7\cdot7}{30\cdot7} - \frac{5\cdot5}{42\cdot5}) = 1 + (\frac{49}{210} - \frac{25}{210}) = 1 + \frac{24}{210} = 1 + \frac{4}{35} = 1\frac{4}{35}$$ 3. a) $$4\frac{5}{6}-2\frac{3}{8}=\frac{29}{6}-\frac{19}{8} = \frac{29 \cdot 4}{6 \cdot 4} - \frac{19 \cdot 3}{8 \cdot 3} = \frac{116}{24} - \frac{57}{24} = \frac{116 - 57}{24} = \frac{59}{24} = 2\frac{11}{24}$$ б) $$5\frac{1}{9}-3\frac{5}{12} = \frac{46}{9} - \frac{41}{12} = \frac{46 \cdot 4}{9 \cdot 4} - \frac{41 \cdot 3}{12 \cdot 3} = \frac{184}{36} - \frac{123}{36} = \frac{184 - 123}{36} = \frac{61}{36} = 1\frac{25}{36}$$