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10 tháng 4 2021

Sửa đề : \(-\dfrac{19}{37}=\dfrac{19}{37}\)

\(\dfrac{19}{37}-\dfrac{31}{24}-\dfrac{19}{37}+1\dfrac{7}{24}-\dfrac{2}{13}\)

\(=\dfrac{19}{37}-\dfrac{31}{24}+\dfrac{19}{37}+\dfrac{31}{24}-\dfrac{2}{13}\)

\(=\left(\dfrac{19}{37}-\dfrac{19}{37}\right)+\left(-\dfrac{31}{24}+\dfrac{31}{24}\right)-\dfrac{2}{13}=-\dfrac{2}{13}\)

 

 

10 tháng 4 2021

ĐỀ ĐÚNG MÀ A

22 tháng 3 2022

\(\left(\dfrac{18}{37}+\dfrac{19}{37}\right)+\left(\dfrac{8}{24}-\dfrac{47}{24}\right)+\dfrac{2}{3}=\dfrac{37}{37}-\dfrac{13}{8}+\dfrac{2}{3}=\dfrac{1}{24}\)

22 tháng 3 2022

ơ hay nhể \(\dfrac{47}{24}\) ở đâu thế

e: \(=\left(\dfrac{18}{37}+\dfrac{19}{37}\right)+\left(\dfrac{8}{24}+\dfrac{2}{3}\right)-\dfrac{47}{24}=2-\dfrac{47}{24}=\dfrac{1}{24}\)

f: \(=-8\cdot\dfrac{1}{2}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)

\(=-4:\dfrac{13}{12}=\dfrac{-48}{13}\)

g: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}-\dfrac{8}{4}=\dfrac{4}{25}+\dfrac{55}{4}-2=\dfrac{1191}{100}\)

giúp mk vs mn ơi , mai cô giáo ktra mk r

a: \(=\dfrac{8}{9}\cdot\dfrac{9}{4}\cdot\dfrac{12}{19}\cdot\dfrac{19}{24}=\dfrac{1}{2}\cdot2=1\)

b: \(=\dfrac{5}{16}\cdot\dfrac{17}{15}\cdot\dfrac{8}{17}=\dfrac{5}{16}\cdot\dfrac{8}{15}=\dfrac{40}{240}=\dfrac{1}{6}\)

c: \(=\dfrac{4}{13}\left(\dfrac{2}{7}+\dfrac{5}{7}\right)-\dfrac{3}{26}=\dfrac{4}{13}-\dfrac{3}{26}=\dfrac{5}{26}\)

c: \(=\dfrac{3}{4}\left(\dfrac{6}{11}+\dfrac{5}{11}\right)-\dfrac{1}{5}=\dfrac{3}{4}-\dfrac{1}{5}=\dfrac{11}{20}\)

26 tháng 2 2022

a, \(\dfrac{14}{13}-\dfrac{1}{13}-\dfrac{19}{20}=1-\dfrac{19}{20}=\dfrac{1}{20}\)

b, \(-\dfrac{24}{17}+\dfrac{7}{17}+\dfrac{1}{16}=\dfrac{-17}{17}+\dfrac{1}{16}=-1+\dfrac{1}{16}=-\dfrac{15}{16}\)

26 tháng 2 2022

\(\dfrac{14}{13}+\left(\dfrac{-1}{13}-\dfrac{19}{20}\right)=\left(\dfrac{14}{13}+\dfrac{-1}{13}\right)-\dfrac{19}{20}=\\ \dfrac{13}{13}-\dfrac{19}{20}=1-\dfrac{19}{20}=\dfrac{20}{20}-\dfrac{19}{20}=\dfrac{1}{20}\)

Câu b em làm tương tự nhé

23 tháng 4 2023

1) \(\dfrac{1}{2}+\dfrac{13}{19}-\dfrac{4}{9}+\dfrac{6}{19}+\dfrac{5}{18}\)

\(=\dfrac{1}{2}+\left(\dfrac{13}{19}+\dfrac{6}{19}\right)-\dfrac{4}{9}+\dfrac{5}{18}\)

\(=\dfrac{3}{2}-\dfrac{4}{9}+\dfrac{5}{18}\)

\(=\dfrac{19}{18}+\dfrac{5}{18}\)

\(=\dfrac{24}{18}\)

\(=\dfrac{4}{3}\)

2) \(\dfrac{-20}{23}+\dfrac{2}{3}-\dfrac{3}{23}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=\left(-\dfrac{20}{23}-\dfrac{3}{23}\right)+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=-1+\dfrac{2}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=-\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{7}{15}\)

\(=\dfrac{1}{15}+\dfrac{7}{15}\)

\(=\dfrac{8}{15}\)

3) \(\dfrac{4}{3}+\dfrac{-11}{31}+\dfrac{3}{10}-\dfrac{20}{31}-\dfrac{2}{5}\)

\(=\left(\dfrac{-11}{31}-\dfrac{20}{31}\right)+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)

\(=-1+\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)

\(=\dfrac{1}{3}+\dfrac{3}{10}-\dfrac{2}{5}\)

\(=\dfrac{1}{3}-\dfrac{1}{10}\)

\(=\dfrac{7}{30}\)

4) \(\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)

\(=\dfrac{5}{7}.\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\)

\(=\dfrac{5}{7}.-\dfrac{7}{11}\)

\(=-\dfrac{35}{77}\)

\(=-\dfrac{5}{11}\)

a: \(=\dfrac{99}{100}:\left(\dfrac{3}{12}-\dfrac{1}{12}+\dfrac{4}{12}\right)-\dfrac{49}{25}\)

\(=\dfrac{99}{100}:\dfrac{1}{2}-\dfrac{49}{25}\)

\(=\dfrac{99}{50}-\dfrac{98}{50}=\dfrac{1}{50}\)

b: \(=\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{32}{60}-1-\dfrac{19}{60}\right):\dfrac{47}{24}\)

\(=\dfrac{39}{60}+\dfrac{-19}{60}\cdot\dfrac{24}{47}\)

=459/940

\(A=\dfrac{636363\cdot37-373737\cdot63}{1+2+3+...+2006}\)

\(=\dfrac{37^2\cdot3^3\cdot7^2\cdot13-37^2\cdot3^3\cdot7^2\cdot13}{\left(2006+1\right)\cdot1003}\)

=0