=18/\(\left(\dfrac{11}{36}+\dfrac{7}{36}-\dfrac{4}{9}\right)\)

\(=18\)\(\)/\(\dfrac{1}{18}=324\)

Bài 1:

\(\dfrac{-13}{38}\) và \(\dfrac{29}{-88}\) 

\(\dfrac{-13}{38}=\dfrac{-13.29}{38.29}=\dfrac{-377}{1102}\) 

\(\dfrac{29}{-88}=\dfrac{-29}{88}=\dfrac{-29.13}{88.13}=\dfrac{-377}{1144}\) 

Vì \(\dfrac{-377}{1102}< \dfrac{-377}{1144}\) nên \(\dfrac{-13}{38}< \dfrac{29}{-88}\) 

\(\dfrac{-18}{31}\) và \(\dfrac{-1818}{3131}\) 

\(\dfrac{-18}{31}\) 

\(\dfrac{-1818}{3131}=\dfrac{-1818:101}{3131:101}=\dfrac{-18}{31}\) 

Vì \(\dfrac{-18}{31}=\dfrac{-18}{31}\) nên \(\dfrac{-18}{31}=\dfrac{-1818}{3131}\)

Bài 2:

a) \(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-4+-3}{156}=\dfrac{-7}{156}\) 

b) \(\dfrac{-6}{9}+\dfrac{-12}{16}=\dfrac{-2}{3}+\dfrac{-3}{4}=\dfrac{-8}{12}+\dfrac{-9}{12}=\dfrac{-17}{12}\)

Vẽ sơ đồ đầu tiên đã

mk giải cho bn r mak ?

B=70.(13.1010156.10101+13.1010172.10101+13.1010190.10101)(56.1010113.10101​+72.1010113.10101​+90.1010113.10101​)

B=70.(1356+1372+1390)(5613​+7213​+9013​)

B=70.39707039​=39

nhớ tick cho mình nhe

cho mình nghi lại nha mn                                                                                                        \(\dfrac{-3}{5}+\dfrac{28}{5}\cdot\left(\dfrac{43}{56}+\dfrac{5}{24}-\dfrac{21}{63}\right)\)

y - 27/54 - 1212/3636 = 1/6

y - 1/2 - 1/3 = 1/6

y = 1/6 + 1/2 + 1/3

y = 1

b: Ta có: \(B=\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+...+\dfrac{1}{64\cdot69}\)

\(=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{64\cdot69}\right)\)

\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{64}-\dfrac{1}{69}\right)\)

\(=\dfrac{1}{5}\cdot\dfrac{65}{4\cdot69}\)

\(=\dfrac{13}{276}\)

\(A=\dfrac{2}{1\cdot4}+\dfrac{2}{4\cdot7}+...+\dfrac{2}{97\cdot100}\\ A=\dfrac{2}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{97\cdot100}\right)\\ A=\dfrac{2}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\\ A=\dfrac{2}{3}\left(1-\dfrac{1}{100}\right)=\dfrac{2}{3}\cdot\dfrac{99}{100}=\dfrac{33}{50}\\ B=\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+...+\dfrac{1}{64\cdot69}\\ B=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{64\cdot69}\right)\\ B=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{64}-\dfrac{1}{69}\right)\\ B=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{69}\right)=\dfrac{1}{5}\cdot\dfrac{65}{276}=\dfrac{13}{276}\)

\(C=70\left(\dfrac{13}{56}+\dfrac{13}{72}+\dfrac{13}{90}\right)=70\cdot13\left(\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\\ C=910\left(\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\\ C=910\left(\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ C=910\left(\dfrac{1}{7}-\dfrac{1}{10}\right)=910\cdot\dfrac{3}{70}=39\)

\(=\dfrac{7}{12}-\dfrac{3}{14}=\dfrac{98-36}{168}=\dfrac{62}{168}=\dfrac{31}{84}\)