Question

A soma entre as frações: a)33/4+8/12 b)4/6+6/18+2/12 c)33/4+8/12+4/6 d)4/5+4/15+2/10 e)33/4+4/6+1/8 f)3/4+4/9 g)4/6+4/9+1/3 h)2/100+4/50 i)3/4+4/6+1/8+3/16 j)2/70+4/140 Os humildes respondem <3

Answer 1

a) $\dfrac{33}{4}+\dfrac{8}{12}$

mmc(4, 12) = 12

$\dfrac{33}{4}+\dfrac{8}{12}=\dfrac{33\cdot3+8}{12}=\dfrac{99+8}{12}=\dfrac{107}{12}$

b) $\dfrac{4}{6}+\dfrac{6}{18}+\dfrac{2}{12}$

mmc(6, 18, 12) = 36

$\dfrac{4}{6}+\dfrac{6}{18}+\dfrac{2}{12}=\dfrac{4\cdot6+6\cdot2+2\cdot3}{36}=\dfrac{24+12+6}{36}$

$\dfrac{4}{6}+\dfrac{6}{18}+\dfrac{2}{12}=\dfrac{42}{36}=\dfrac{7}{6}$

c) $\dfrac{33}{4}+\dfrac{8}{12}+\dfrac{4}{6}$

mmc(4, 12, 6) = 12

$\dfrac{33}{4}+\dfrac{8}{12}+\dfrac{4}{6}=\dfrac{33\cdot3+8\cdot1+4\cdot2}{12}=\dfrac{99+8+8}{12}$

$\dfrac{33}{4}+\dfrac{8}{12}+\dfrac{4}{6}=\dfrac{115}{12}$

d) $\dfrac{4}{5}+\dfrac{4}{15}+\dfrac{2}{10}$

mmc(5, 10, 15) = 30

$\dfrac{4}{5}+\dfrac{4}{15}+\dfrac{2}{10}=\dfrac{4\cdot6+4\cdot2+2\cdot3}{30}=\dfrac{24+8+6}{30}$

$\dfrac{4}{5}+\dfrac{4}{15}+\dfrac{2}{10}=\dfrac{38}{30}=\dfrac{19}{15}$

e) $\dfrac{33}{4}+\dfrac{4}{6}+\dfrac{1}{8}$

mmc(4, 6, 8) = 24

$\dfrac{33}{4}+\dfrac{4}{6}+\dfrac{1}{8}=\dfrac{33\cdot6+4\cdot4+1\cdot3}{24}=\dfrac{198+16+3}{24}$

$\dfrac{33}{4}+\dfrac{4}{6}+\dfrac{1}{8}=\dfrac{217}{24}$

f) $\dfrac{3}{4}+\dfrac{4}{9}$

mmc(4, 9) = 36

$\dfrac{3}{4}+\dfrac{4}{9}=\dfrac{3\cdot9+4\cdot4}{36}=\dfrac{27+16}{36}$

$\dfrac{3}{4}+\dfrac{4}{9}=\dfrac{43}{36}$

g) $\dfrac{4}{6}+\dfrac{4}{9}+\dfrac{1}{3}$

mmc(6, 9, 3) = 18

$\dfrac{4}{6}+\dfrac{4}{9}+\dfrac{1}{3}=\dfrac{4\cdot3+4\cdot2+1\cdot6}{18}=\dfrac{12+8+6}{18}$

$\dfrac{4}{6}+\dfrac{4}{9}+\dfrac{1}{3}=\dfrac{26}{18}=\dfrac{13}{9}$

h) $\dfrac{2}{100}+\dfrac{4}{50}$

mmc(100, 50) = 100

$\dfrac{2}{100}+\dfrac{4}{50}=\dfrac{2\cdot1+4\cdot2}{100}=\dfrac{2+8}{100}$

$\dfrac{2}{100}+\dfrac{4}{50}=\dfrac{10}{100}=\dfrac{1}{10}$

i) $\dfrac{3}{4}+\dfrac{4}{6}+\dfrac{1}{8}+\dfrac{3}{16}$

mmc(4, 6, 8, 16) = 48

$\dfrac{3}{4}+\dfrac{4}{6}+\dfrac{1}{8}+\dfrac{3}{16}=\dfrac{3\cdot12+4\cdot8+1\cdot6+3\cdot3}{48}$

$\dfrac{3}{4}+\dfrac{4}{6}+\dfrac{1}{8}+\dfrac{3}{16}=\dfrac{36+32+6+9}{48}=\dfrac{83}{48}$

j) $\dfrac{2}{70}+\dfrac{4}{140}$

mmc(70, 140) = 140

$\dfrac{2}{70}+\dfrac{4}{140}=\dfrac{2\cdot2+4\cdot1}{140}=\dfrac{4+4}{140}$

$\dfrac{2}{70}+\dfrac{4}{140}=\dfrac{8}{140}=\dfrac{2}{35}$