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1.
a) \(\frac{16}{24}-\frac{1}{3}=\frac{16}{24}-\frac{8}{24}=\)\(\frac{8}{24}=\frac{1}{3}\)
b) \(\frac{4}{5}-\frac{12}{60}=\frac{48}{60}-\frac{12}{60}=\frac{36}{60}=\frac{9}{15}\)
3.
a)\(\frac{17}{6}-\frac{2}{6}=\frac{17-2}{6}=\frac{15}{6}\)
b) \(\frac{16}{15}-\frac{11}{15}=\frac{16-11}{15}=\frac{5}{15}=\frac{1}{3}\)
c) \(\frac{19}{12}-\frac{13}{12}=\frac{19-13}{12}=\frac{6}{12}=\frac{1}{2}\)
a,\(\frac{26}{7}\)
b,\(\frac{78}{19}\)
c,\(\frac{13}{11}\)
d,\(1\)
a, \(4-\frac{2}{7}\)=\(\frac{26}{7}\)
b,5-\(\frac{17}{19}=\frac{78}{19}\)
c,\(\frac{35}{11}-2=\frac{13}{11}\)
d,\(\frac{68}{17}-3=1\)
k mình nha
a) \(\frac{13}{19}+\frac{18}{19}+\frac{19}{19}=\left(\frac{13}{19}+\frac{17}{19}\right)+\frac{18}{19}=\frac{20}{19}+\frac{18}{19}=\frac{48}{19}\)
b) \(\frac{3}{5}+\frac{3}{16}+\frac{13}{16}=\left(\frac{3}{16}+\frac{13}{16}\right)+\frac{3}{5}=1+\frac{3}{5}=\frac{5}{5}+\frac{3}{5}=\frac{8}{5}\)
a) \(\frac{13}{19}+\frac{18}{19}+\frac{17}{19}\)
= \(\frac{31}{19}+\frac{17}{19}\)
= \(\frac{48}{19}.\)
b) \(\frac{3}{5}+\frac{3}{16}+\frac{13}{16}\)
= \(\frac{3}{5}+\left(\frac{3}{16}+\frac{13}{16}\right)\)
= \(\frac{3}{5}+\frac{16}{16}\)
= \(\frac{3}{5}+1\)
= \(\frac{3}{5}.\)
a) \(\frac{2}{5}+\frac{9}{15}=\frac{2}{5}+\frac{3}{5}=\frac{5}{5}=1\)
b) \(\frac{15}{45}+\frac{25}{30}=\frac{1}{3}+\frac{5}{6}=\frac{2}{6}+\frac{5}{6}=\frac{7}{6}\)
c) \(\frac{28}{32}+\frac{45}{72}=\frac{7}{8}+\frac{5}{8}=\frac{12}{8}=\frac{3}{2}\)
d) \(\frac{8}{28}+\frac{5}{30}=\frac{2}{7}+\frac{1}{6}=\frac{12}{42}+\frac{7}{42}=\frac{19}{42}\)
a) \(\frac{2}{5}+\frac{9}{15}\)=\(\frac{2}{5}+\frac{3}{5}=\frac{2+3}{5}=1\)
b)\(\frac{15}{45}+\frac{25}{30}=\frac{1}{3}+\frac{5}{6}=\frac{2}{6}+\frac{5}{6}=\frac{2+5}{6}=\frac{7}{6}\)
c)\(\frac{28}{32}+\frac{45}{72}=\frac{7}{8}+\frac{5}{8}=\frac{7+5}{8}=\frac{12}{8}=\frac{3}{2}\)
d)\(\frac{8}{28}+\frac{5}{30}=\frac{2}{7}+\frac{1}{6}=\frac{12}{42}+\frac{7}{42}=\frac{12+7}{42}=\frac{19}{42}\)
a) \(3\frac{4}{7}\times2\frac{4}{5}\)
\(=\frac{25}{7}\times\frac{14}{5}=\frac{350}{35}=10\)
b) \(\frac{4}{5}:\frac{1}{2}+\frac{17}{3}\)
\(=\frac{8}{5}+\frac{17}{3}\)
\(=\frac{109}{15}\)
c) \(5\frac{1}{3}-2\frac{1}{6}\)
\(=\frac{16}{3}-\frac{13}{6}\)
\(=\frac{32}{6}-\frac{13}{6}=\frac{19}{6}\)
d) \(4\frac{1}{2}:2\frac{1}{4}\)
\(=\frac{9}{2}:\frac{9}{4}=\frac{45}{18}\)
Ta thấy 2700 là có 2 số 0 thì phải nhân cho 100
=> ( ab + 13) = 100 => ab = 87
Vậy ta tìm được a=8 và b=7 thỏa mãn điều kiện a-b =1
=> 100 x cd = 2700 => cd= \(\frac{2700}{100}\)= 27
Vậy ta tìm được c=2 và d=7
Vậy a=8, b=7 ,c =2 và d=7
Còn ( ba + 13) x cd thì a=3, b= 2, c= 7 và d=5
\(a,\frac{10}{21}×\frac{14}{25}=\frac{4}{15}\)
\(b,3\frac{1}{3}×\frac{12}{35}=\frac{10}{3}×\frac{12}{35}=\frac{8}{7}\)
\(c,\frac{7}{15}×\frac{11}{13}+\frac{7}{15}×\frac{2}{13}+\frac{8}{15}=\frac{7}{15}+\frac{8}{15}=1\)
\(d,\left(\frac{41}{75}+\frac{17}{10}\right)×\frac{80}{129}=\frac{4}{9}\)
a) 11162096
b) 470546
c) 50158
d) 351