\(\Leftrightarrow D=1-\dfrac{1}{3}-\dfrac{1}{6}-\dfrac{1}{10}-\dfrac{1}{15}-\dfrac{1}{21}-\dfrac{1}{28}\)

\(\Rightarrow\dfrac{1}{2}D=\dfrac{1}{2}-\dfrac{1}{2.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}-\dfrac{1}{5.6}-\dfrac{1}{6.7}-\dfrac{1}{7.8}\)

\(\Rightarrow D\dfrac{1}{2}=\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{7}+\dfrac{1}{8}\)

\(\Rightarrow D=\dfrac{1}{8}.2=\dfrac{1}{4}\)

Vậy D=1/4

Ta có:

\(\dfrac{1}{3}\times\dfrac{12}{12}=\dfrac{12}{36};\)

\(\dfrac{1}{6}\times\dfrac{6}{6}=\dfrac{6}{36};\)

\(\dfrac{1}{10}\times\dfrac{3}{3}=\dfrac{3}{30};\)

\(\dfrac{1}{15}\times\dfrac{2}{2}=\dfrac{2}{30};\)

\(\dfrac{1}{21}\times\dfrac{4}{4}=\dfrac{4}{84};\)

\(\dfrac{1}{28}\times\dfrac{3}{3}=\dfrac{3}{84};\)

\(A=\dfrac{12}{36}+\dfrac{6}{36}+\dfrac{3}{30}+\dfrac{2}{30}+\dfrac{4}{84}+\dfrac{3}{84}+\dfrac{1}{36}\)

    \(=\left(\dfrac{12}{36}+\dfrac{6}{36}+\dfrac{1}{36}\right)+\left(\dfrac{3}{30}+\dfrac{2}{30}\right)+\left(\dfrac{4}{84}+\dfrac{3}{84}\right)\)

    \(=\dfrac{19}{36}+\dfrac{5}{30}+\dfrac{7}{84}\)

    \(=\dfrac{19}{36}+\dfrac{1}{6}+\dfrac{1}{12}\)

    \(=\dfrac{19}{36}+\dfrac{6}{36}+\dfrac{3}{36}\)

    \(=\dfrac{28}{36}=\dfrac{7}{9}\)

Vậy: \(A=\dfrac{7}{9}\)

a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)

=\(\dfrac{10}{11}.\dfrac{-8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)

=\(\dfrac{10}{11}(\dfrac{-8}{9}+\dfrac{7}{18})\)

=\(\dfrac{10}{11}.\dfrac{-1}{2}\)

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

b; 

B = \(\dfrac{3}{14}\) : \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\) : \(\dfrac{1}{28}\) - 8

B = (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) - 8

B = (\(\dfrac{9}{42}\) - \(\dfrac{26}{42}\) + \(\dfrac{29}{42}\)) - 8

B = (\(\dfrac{-17}{42}\) + \(\dfrac{29}{42}\)) - 8

B = \(\dfrac{2}{7}\) - 8

B = \(\dfrac{2}{7}-\dfrac{56}{7}\)

B = - \(\dfrac{54}{7}\)

c; C = -1\(\dfrac{5}{7}\).15 + \(\dfrac{2}{7}\)(-15) + (-105).(\(\dfrac{2}{3}\) - \(\dfrac{4}{5}\) + \(\dfrac{1}{7}\))

C = - 15.(- 1 - \(\dfrac{5}{7}\) + \(\dfrac{2}{7}\) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\) + \(1\))

C = -15.[(1 - 1) - (\(\dfrac{5}{7}\) - \(\dfrac{2}{7}\)) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\)]

C = -15.[0 - \(\dfrac{3}{7}\) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\)]

C = -15 . [- \(\dfrac{45}{105}\) + \(\dfrac{490}{105}\)  - \(\dfrac{588}{105}\)]

C = -15. [ \(\dfrac{445}{105}\) - \(\dfrac{588}{105}\)]

C = - 15.(- \(\dfrac{143}{105}\))

C = \(\dfrac{143}{7}\)

\(F=\dfrac{5}{6}+6\dfrac{5}{6}\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)

\(F=\dfrac{5}{6}+\dfrac{41}{6}\left(\dfrac{225}{20}-\dfrac{37}{4}\right):\dfrac{25}{3}\)

\(F=\dfrac{5}{6}+\dfrac{41}{6}.2.\dfrac{3}{25}\)

\(F=\dfrac{5}{6}+\dfrac{41}{25}.\dfrac{3}{25}\)

\(F=\dfrac{5}{6}+\dfrac{41}{25}\)

\(F=\dfrac{371}{150}\)

\(D=\left(\dfrac{136}{15}-\dfrac{28}{5}+\dfrac{62}{10}\right)\times\dfrac{21}{24}\)

\(D=\left(\dfrac{272}{30}-\dfrac{168}{30}+\dfrac{186}{30}\right)\times\dfrac{21}{24}\)

\(D=\dfrac{290}{30}\times\dfrac{21}{24}\)

\(D=\dfrac{29}{3}\times\dfrac{7}{8}\)

\(D=\dfrac{203}{24}\)

a=78/35

b=22/12

c=1/1

d=40202090/4040090

e=1,24025667172...

f=871,82

ko biết đúng ko [0_0'] hihi

A =\(2.\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+......+\dfrac{1}{156}\right)\)

A =\(2.\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+..........+\dfrac{1}{12.13}\right)\)

A =2.\(\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)

A=\(2.\dfrac{10}{39}=\dfrac{20}{39}\)

tớ làm hơi gọn nên có gì kho hiểu thì nói tớ

a) \(\dfrac{4}{24}+\dfrac{7}{6}=\dfrac{4}{24}+\dfrac{28}{24}=\dfrac{4+28}{24}=\dfrac{32}{24}=\dfrac{4}{3}\)

b) \(\dfrac{10}{15}-\dfrac{1}{3}=\dfrac{10}{15}-\dfrac{5}{15}=\dfrac{10-5}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)

c) \(\dfrac{21}{28}-\dfrac{1}{4}=\dfrac{21}{28}-\dfrac{7}{28}=\dfrac{21-7}{28}=\dfrac{14}{28}=\dfrac{1}{2}\)

d) \(\dfrac{35}{40}+\dfrac{5}{8}=\dfrac{35}{40}+\dfrac{25}{40}=\dfrac{35+25}{40}=\dfrac{60}{40}=\dfrac{3}{2}\)

\(a)\dfrac{4}{24}=\dfrac{1}{6} \\ \dfrac{1}{6}+\dfrac{7}{6}\\ =\dfrac{8}{6}=\dfrac{4}{3}\\ b)\dfrac{10}{15}=\dfrac{2}{3}-\dfrac{1}{3}\\ =\dfrac{1}{3}\\ c)\dfrac{21}{28}=\dfrac{3}{4}\\ \dfrac{3}{4}-\dfrac{1}{4}\\ =\dfrac{2}{4}=\dfrac{1}{2}\\ d)\dfrac{35}{40}=\dfrac{7}{8}\\ \dfrac{7}{8}+\dfrac{5}{8}\\ =\dfrac{12}{8}=\dfrac{3}{2}\)

\(M=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{105}+\dfrac{1}{120}\)

\(M=2.\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+...+\dfrac{1}{240}\right)\)

\(M=2.\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{15.16}\right)\)

\(M=2.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

\(M=2.\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\)

\(M=2.\dfrac{3}{16}\)

\(M=\dfrac{3}{8}\)

Vậy \(\dfrac{1}{3}< M< \dfrac{1}{2}\)