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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: \(A=\dfrac{7}{9}\)

    

10 tháng 4 2017

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

10 tháng 4 2017

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

30 tháng 3 2017

=\(\dfrac{1}{3.2}+\dfrac{1}{2.5}+\dfrac{1}{5.3}+\dfrac{1}{3.7}+\dfrac{1}{7.4}+\dfrac{1}{4.9}+\dfrac{1}{9.5}\)=\(\dfrac{1}{3}+\dfrac{1}{5}\)

=\(\dfrac{8}{15}\)
30 tháng 3 2017

Gọi A = \(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}\)

\(\dfrac{1}{2}\)A = \(\dfrac{1}{2}.\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}\right)\)

\(\dfrac{1}{2}\)A = \(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)

\(\dfrac{1}{2}\)A = \(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

\(\dfrac{1}{2}\)A = \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

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

\(\dfrac{1}{2}\)A = \(\dfrac{7}{30}\)

A = \(\dfrac{7}{30}:\dfrac{1}{2}\)

A = \(\dfrac{7}{15}\)

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

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

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

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

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

7 tháng 8

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 = - \(\dfrac{54}{7}\)

27 tháng 7 2021

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

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

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

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

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

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

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

Vậy \(\dfrac{1}{3}< M< \dfrac{1}{2}\)

1 tháng 3 2023

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

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

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

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

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

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

1 tháng 3 2023

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

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

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

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

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

29 tháng 4 2018

\(\dfrac{1}{2}N=\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)

\(\dfrac{1}{2}N=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

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

N=\(\dfrac{2}{5}:\dfrac{1}{2}=\dfrac{4}{5}\)

14 tháng 3 2017

Ta có: \(A=\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{6}-1\right)\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{15}-1\right)\left(\dfrac{1}{21}-1\right)\left(\dfrac{1}{28}-1\right)\left(\dfrac{1}{36}-1\right)\)

\(=\dfrac{-2}{3}.\dfrac{-5}{6}.\dfrac{-9}{10}.\dfrac{-14}{15}.\dfrac{-20}{21}.\dfrac{-27}{28}.\dfrac{-35}{36}\)

\(=\dfrac{-2.\left(-5\right).3.\left(-3\right).2.\left(-7\right).\left(-4\right).5.\left(-3\right).9.5.\left(-7\right)}{3.2.3.2.5.3.5.3.7.4.7.4.9}\)

\(=\dfrac{-5}{3.4}=\dfrac{-5}{12}\)

Vậy \(A=\dfrac{-5}{12}.\)

14 tháng 3 2017

\(C=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\)

\(2C=2\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\right)\)

\(2C=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{2015}}\)

\(2C-C=\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2015}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2016}}\right)\)

\(C=2-\dfrac{1}{2^{2016}}\)

21 tháng 7 2017

+) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)

\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)

\(\Rightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)

\(\Rightarrow A=1-\dfrac{1}{2^{10}}=\dfrac{2^{10}-1}{2^{10}}\)

Vậy \(A=\dfrac{2^{10}-1}{2^{10}}\)

+) \(F=\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{1}{190}\)

\(\Rightarrow\dfrac{1}{2}F=\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{380}\)

\(=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{19.20}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{20}\)

\(=\dfrac{1}{5}-\dfrac{1}{20}=\dfrac{3}{20}\Rightarrow F=\dfrac{3}{20}:\dfrac{1}{2}=\dfrac{3}{10}\)

Vậy \(F=\dfrac{3}{10}\)

+) \(G=\dfrac{12}{84}+\dfrac{12}{210}+\dfrac{12}{390}+...+\dfrac{12}{2100}\)

\(=\dfrac{4}{28}+\dfrac{4}{70}+\dfrac{4}{130}+...+\dfrac{4}{700}=\dfrac{4}{4.7}+\dfrac{4}{7.10}+...+\dfrac{4}{25.28}\)

\(=\dfrac{4}{3}.\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{25.28}\right)\)

\(=\dfrac{4}{3}.\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)

\(=\dfrac{4}{3}.\left(\dfrac{1}{4}-\dfrac{1}{28}\right)=\dfrac{4}{3}.\dfrac{3}{14}=\dfrac{2}{7}\)

Vậy \(G=\dfrac{2}{7}\)

21 tháng 7 2017

\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)

\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)

\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\right)\)

\(A=1-\dfrac{1}{2^{10}}=\dfrac{1024-1}{1024}=\dfrac{1023}{1024}\)

\(F=\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{1}{190}\)

\(=\dfrac{2}{30}+\dfrac{2}{42}+\dfrac{2}{56}+...+\dfrac{2}{380}\)

\(=\dfrac{2}{5.6}+\dfrac{2}{6.7}+\dfrac{2}{7.8}+...+\dfrac{2}{19.20}\)

\(=2\left(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{19.20}\right)\)

\(=2\left(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)

\(=2\left(\dfrac{1}{5}-\dfrac{1}{20}\right)=2.\dfrac{3}{20}=\dfrac{3}{10}\)

\(G=\dfrac{12}{84}+\dfrac{12}{210}+\dfrac{12}{390}+...+\dfrac{12}{2100}\)

\(=\dfrac{4}{28}+\dfrac{4}{70}+\dfrac{4}{130}+...+\dfrac{4}{700}\)

\(=\dfrac{4}{4.7}+\dfrac{4}{7.10}+\dfrac{4}{10.13}+...+\dfrac{4}{25.28}\)

\(=\dfrac{4}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)

\(=\dfrac{4}{3}\left(\dfrac{1}{4}-\dfrac{1}{28}\right)\)

\(=\dfrac{4}{3}.\dfrac{3}{14}=\dfrac{2}{7}\)