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24 tháng 3 2016

Đặt A = 1/10 + 1/15 + 1/21 + ... + 1/66

=> 1/2.A = 1/20 + 1/30 + 1/42 + ... + 1/132

=> 1/2.A = 1/4.5 + 1/5.6 + 1/6.7 + ... + 1/11.12

=> 1/2.A = 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + ... + 1/11 - 1/12

=> 1/2.A = 1/4 - 1/12

=> 1/2.A = 1/6

=> A = 1/6 : 1/2

=> A = 1/3

Vậy 1/10 + 1/15 + ... + 1/66 = 1/3

24 tháng 3 2016

Đặt A = 1/10 + 1/15 + 1/21 + ... + 1/66

=> 1/2.A = 1/20 + 1/30 + 1/42 + ... + 1/132

=> 1/2.A = 1/4.5 + 1/5.6 + 1/6.7 + ... + 1/11.12

=> 1/2.A = 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + ... + 1/11 - 1/12

=> 1/2.A = 1/4 - 1/12

=> 1/2.A = 1/6

=> A = 1/6 : 1/2

=> A = 1/3

Vậy 1/10 + 1/15 + ... + 1/66 = 1/3

27 tháng 4 2016

\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{66}\)

\(A=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{132}\)

\(A=\frac{1}{2}.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{132}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{4}-\frac{1}{12}\right)\)

\(A=\frac{1}{2}.\frac{1}{6}\)

\(A=\frac{1}{12}\)

8 tháng 8 2020

Bài làm:

Ta có: \(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{66}\)

\(=\frac{1}{1}+\frac{1}{1.3}+\frac{1}{3.2}+...+\frac{1}{11.6}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.1.3}+\frac{1}{2.3.2}+...+\frac{1}{2.11.6}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{11.12}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{12}\right)\)

\(=\frac{1}{2}.\frac{11}{12}\)

\(=\frac{11}{24}\)

8 tháng 8 2020

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}+\frac{1}{66}\)

\(=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{90}+\frac{2}{110}+\frac{2}{132}\)

\(=2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{9\times10}+\frac{1}{10\times11}+\frac{1}{11\times12}\right)\)

\(=2\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\right)\)

\(=2\times\left(1-\frac{1}{12}\right)\)

\(=2\times\frac{11}{12}\)

\(=\frac{11}{6}\)

3 tháng 4 2016

\(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+........+\frac{1}{66}\)

=\(\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...........+\frac{2}{132}\)

=\(2\left(\frac{1}{4.5}+\frac{1}{5.6}+..........+\frac{1}{11.12}\right)\)

=\(2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+..........+\frac{1}{11}-\frac{1}{12}\right)\)

=\(2\left(\frac{1}{4}-\frac{1}{12}\right)\)

=\(2.\frac{1}{6}\)

=\(\frac{1}{3}\)

18 tháng 4 2019

\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{66}\)

\(\frac{A}{2}=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{132}\)

\(\frac{A}{2}=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{11\cdot12}\)

\(\frac{A}{2}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{11}-\frac{1}{12}\)

\(\frac{A}{2}=\frac{1}{4}-\frac{1}{12}\)

\(\Rightarrow A=\frac{2}{4}-\frac{2}{12}=\frac{16}{48}\)

\(B=\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{55}\)

\(\frac{B}{2}=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{110}\)

\(\frac{B}{2}=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{10\cdot11}\)

\(\frac{B}{2}=\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)

\(\frac{B}{2}=\frac{1}{3}-\frac{1}{11}\)

\(\Rightarrow B=\frac{2}{3}-\frac{2}{11}=\frac{16}{33}\)

Mà \(\frac{16}{48}< \frac{16}{33}\Rightarrow A< B\)

Vậy : A < B

6 tháng 5 2015

a)\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{195}\)

Đặt \(C=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{66}\)

\(\Rightarrow\frac{1}{2}C=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{132}\)

\(\Rightarrow\frac{1}{2}C=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\)

\(\Rightarrow\frac{1}{2}C=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)

\(\Rightarrow\frac{1}{2}C=\frac{1}{4}+\left(-\frac{1}{5}+\frac{1}{5}\right)+\left(-\frac{1}{6}+\frac{1}{6}\right)+...+\left(-\frac{1}{11}+\frac{1}{11}\right)-\frac{1}{12}\)\(\Rightarrow\frac{1}{2}C=\frac{1}{4}+0+0+...+0-\frac{1}{12}\)

\(\Rightarrow\frac{1}{2}C=\frac{1}{4}-\frac{1}{12}\)

\(\Rightarrow\frac{1}{2}C=\frac{3}{12}-\frac{1}{12}\)

\(\Rightarrow\frac{1}{2}C=\frac{2}{12}\)

\(\Rightarrow\frac{1}{2}C=\frac{1}{6}\)

\(\Rightarrow C=\frac{1}{6}:\frac{1}{2}\)

\(\Rightarrow C=\frac{1}{6}\cdot2\)

\(\Rightarrow C=\frac{2}{6}=\frac{1}{3}\)