1) \(12\frac{3}{8}+8\frac{1}{6}=20+\frac{9}{24}+\frac{4}{24}=20\frac{13}{24}\).
2) \(3\frac{13}{15}+6\frac{7}{10}=9+\frac{26}{30}+\frac{21}{30}=9\frac{47}{30}=10\frac{17}{30}\).
3) \(9\frac{8}{21}+4\frac{11}{14}=13+\frac{16}{42}+\frac{33}{42}=13\frac{49}{42}=14\frac{1}{6}\).
4) \(8\frac{5}{9}+7\frac{3}{4}+12\frac{7}{12}=27+\frac{20}{36}+\frac{27}{36}+\frac{21}{36}=28\frac{17}{18}\).
Ответ: 1) \(20\frac{13}{24}\); 2) \(10\frac{17}{30}\); 3) \(14\frac{1}{6}\); 4) \(28\frac{17}{18}\).
1) \(1-\frac{7}{15}=\frac{8}{15}\); 2) \(3-\frac{9}{23}=2\frac{14}{23}\); 3) \(4-1\frac{2}{3}=2\frac{1}{3}\); 4) \(10-5\frac{4}{11}=4\frac{7}{11}\).
Ответ: 1) \(\frac{8}{15}\); 2) \(2\frac{14}{23}\); 3) \(2\frac{1}{3}\); 4) \(4\frac{7}{11}\).
1) \(7\frac{5}{6}-3\frac{2}{3}=4\frac{1}{6}\); 2) \(8\frac{5}{12}-6\frac{7}{20}=1\frac{4}{15}\); 3) \(11\frac{11}{12}-5\frac{7}{9}=6\frac{5}{36}\); 4) \(9\frac{17}{24}-8\frac{11}{36}=1\frac{7}{72}\).
Ответ: 1) \(4\frac{1}{6}\); 2) \(1\frac{4}{15}\); 3) \(6\frac{5}{36}\); 4) \(1\frac{7}{72}\).
1) \(3\frac{1}{16}-\frac{1}{8}=2\frac{15}{16}\); 2) \(7\frac{9}{20}-5\frac{17}{30}=1\frac{11}{60}\); 3) \(4\frac{2}{7}-1\frac{4}{9}=2\frac{41}{63}\); 4) \(8\frac{5}{36}-1\frac{43}{108}=6\frac{7}{108}\); 5) \(9\frac{7}{9}-4\frac{5}{6}=4\frac{17}{18}\); 6) \(6\frac{7}{32}-2\frac{11}{48}=4\frac{5}{24}\).
Ответ: 1) \(2\frac{15}{16}\); 2) \(1\frac{11}{60}\); 3) \(2\frac{41}{63}\); 4) \(6\frac{7}{108}\); 5) \(4\frac{17}{18}\); 6) \(4\frac{5}{24}\).
1) \(x=10\frac{5}{8}-7\frac{3}{5}=3\frac{1}{40}\).
2) \(x-2\frac{7}{8}=4\frac{2}{3}-3\frac{5}{6}=\frac{5}{6}\), поэтому \(x=2\frac{7}{8}+\frac{5}{6}=3\frac{17}{24}\).
Ответ: 1) \(x=3\frac{1}{40}\); 2) \(x=3\frac{17}{24}\).
1) \(6\frac{7}{8}-3\frac{1}{3}+5\frac{5}{16}=8\frac{53}{48}=9\frac{5}{48}\).
2) \(5\frac{9}{14}-2\frac{3}{7}+6.7=9\frac{13}{14}\).
3) \((15\frac{5}{6}-9\frac{25}{27})-2\frac{17}{18}=3\frac{17}{27}\).
4) \((18-10\frac{5}{9})-(6\frac{1}{8}-3\frac{2}{3})=5\frac{67}{72}\).
Ответ: 1) \(9\frac{5}{48}\); 2) \(9\frac{13}{14}\); 3) \(3\frac{17}{27}\); 4) \(5\frac{67}{72}\).