a) 1/10 + 1/15 + 1/21 +.....+ 1/66
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LP
1
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
PT
1
27 tháng 4 2016
\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{66}\)
\(A=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{132}\)
\(A=\frac{1}{2}.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{132}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{4}-\frac{1}{12}\right)\)
\(A=\frac{1}{2}.\frac{1}{6}\)
\(A=\frac{1}{12}\)
KB
3 tháng 4 2016
\(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+........+\frac{1}{66}\)
=\(\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...........+\frac{2}{132}\)
=\(2\left(\frac{1}{4.5}+\frac{1}{5.6}+..........+\frac{1}{11.12}\right)\)
=\(2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+..........+\frac{1}{11}-\frac{1}{12}\right)\)
=\(2\left(\frac{1}{4}-\frac{1}{12}\right)\)
=\(2.\frac{1}{6}\)
=\(\frac{1}{3}\)
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
LH
1
NP
6 tháng 5 2015
a)\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{195}\)
Đặt \(C=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{66}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{132}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{4}+\left(-\frac{1}{5}+\frac{1}{5}\right)+\left(-\frac{1}{6}+\frac{1}{6}\right)+...+\left(-\frac{1}{11}+\frac{1}{11}\right)-\frac{1}{12}\)\(\Rightarrow\frac{1}{2}C=\frac{1}{4}+0+0+...+0-\frac{1}{12}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{4}-\frac{1}{12}\)
\(\Rightarrow\frac{1}{2}C=\frac{3}{12}-\frac{1}{12}\)
\(\Rightarrow\frac{1}{2}C=\frac{2}{12}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{6}\)
\(\Rightarrow C=\frac{1}{6}:\frac{1}{2}\)
\(\Rightarrow C=\frac{1}{6}\cdot2\)
\(\Rightarrow C=\frac{2}{6}=\frac{1}{3}\)
Q
30 tháng 3 2017
\(A=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{55}+\dfrac{1}{66}\)
\(A=2\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\right)\)
\(A=2\left(\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}\right)\)
\(A=2\left(\left(\dfrac{1}{4}-\dfrac{1}{5}\right)+\left(\dfrac{1}{5}-\dfrac{1}{6}\right)+\left(\dfrac{1}{6}-\dfrac{1}{7}\right)+\left(\dfrac{1}{7}-\dfrac{1}{8}\right)+\left(\dfrac{1}{8}-\dfrac{1}{9}\right)+\left(\dfrac{1}{9}-\dfrac{1}{10}\right)+\left(\dfrac{1}{10}-\dfrac{1}{11}\right)+\left(\dfrac{1}{11}-\dfrac{1}{12}\right)\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{12}\right)\Rightarrow A=\dfrac{1}{3}\)
LC
3
8 tháng 8 2020
Bài làm:
Ta có: \(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{66}\)
\(=\frac{1}{1}+\frac{1}{1.3}+\frac{1}{3.2}+...+\frac{1}{11.6}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.1.3}+\frac{1}{2.3.2}+...+\frac{1}{2.11.6}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{11.12}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{12}\right)\)
\(=\frac{1}{2}.\frac{11}{12}\)
\(=\frac{11}{24}\)
NH
8 tháng 8 2020
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}+\frac{1}{66}\)
\(=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{90}+\frac{2}{110}+\frac{2}{132}\)
\(=2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{9\times10}+\frac{1}{10\times11}+\frac{1}{11\times12}\right)\)
\(=2\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\right)\)
\(=2\times\left(1-\frac{1}{12}\right)\)
\(=2\times\frac{11}{12}\)
\(=\frac{11}{6}\)
TB
0
AN
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=\(2.\dfrac{10}{39}=\dfrac{20}{39}\)