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23 tháng 1 2022

2P=2/1.2.3+2/2.3.4+2/3.4.5+2/10.11.12
2P=1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5+.....+1/10.11-1/11.12
2P=1/1.2-1/11.12
2P=1/2-1/132
2P=66/132-1/132
2P=65/132
 P=65/264

\(P=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{10.11.12}\)

\(P=\dfrac{1}{2}-\dfrac{1}{11.12}\)

\(P=\dfrac{65}{132}\)

 

30 tháng 3 2016

\(M=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+.....+\frac{1}{10.11}-\frac{1}{11.12}\)

\(M=\frac{1}{2}-\frac{1}{11.12}=\frac{65}{132}\)

17 tháng 4 2017

Nhận xét rằng:
2/[(n - 1)n(n +1)] = 1/[(n-1).n] - 1/[n(n+1)]
Do đó
2M = 2/(1.2.3) + 2/(2.3.4) + 2/(3.4.5) + ... + 2(10.11.12)
= 1/(1.2) - 1/(2.3) + 1/(2.3) - 1/(3.4) + 1/(3.4) - 1/(4.5) + .... + 1/(10.11) - 1/(11.12)
= 1/(1.2) - 1/(11.12) = 65/132
=> M = 65/264

17 tháng 4 2017

Ta có nhận xét: \(\dfrac{1}{1.2}-\dfrac{1}{2.3}=\dfrac{3-1}{1.2.3}=\dfrac{2}{1.2.3}\),

\(\dfrac{1}{2.3}-\dfrac{1}{3.4}=\dfrac{4-2}{2.3.4}=\dfrac{2}{2.3.4};...\)

\(\Rightarrow\dfrac{1}{1.2.3}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)\);

\(\dfrac{1}{2.3.4}=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)\);...

Do đó \(M=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{1.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{11.12}\right)=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{11.12}\right)\)

\(=\dfrac{1}{2}.\dfrac{65}{132}=\dfrac{65}{264}\)

\(M=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+.....+\frac{1}{10.11.12}\)

\(M=\frac{1}{2}-\frac{1}{11.12}\)

\(M=\frac{65}{132}\)

Ngắn gọn , xúc tích !!! :))

 
28 tháng 7 2017

\(M=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\right)\)

\(M=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{11.12}\right)\)

\(M=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{132}\right)\)

1 tháng 5 2015

\(M=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\)

\(=\frac{1}{2}-\frac{1}{11.12}\)

\(=\frac{65}{132}\)

1 tháng 5 2015

Ta có nhận xét: 1/1.2 - 1/2.3 = 3-1/1.2.3 = 2/1.2.3          

                       1/2.3 - 1/3.4 = 4-2/2.3.4 = 2/2.3.4      

Suy ra:             1/1.2.3 = 1/2(1/1.2 - 1/2.3)    

                         1/2.3.4 = 1/2(1/2.3 -1/3.4)  

Do đó:              M = 1/2(1/1.2-1/2.3 + 1/2.3 -1/3.4 + ... + 1/10.11 -1/11.12)

                           = 1/2(1/1.2 - 1/11.12) = 1/2(1/2-11/12 )      

                           = 1/2.65/132 = 65/264

Phức tạp lắm

6 tháng 3 2019

\(D=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{10\cdot11\cdot12}\)

\(D=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{10\cdot11\cdot12}\right)\)

\(D=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{10\cdot11}-\frac{1}{11\cdot12}\right)\)

\(D=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{132}\right)=...\)

6 tháng 3 2019

\(D=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{10.11.12}\)

\(D=\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{10.11.12}\right).\frac{1}{2}\)

\(D=\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{10.11}-\frac{1}{11.12}\right).\frac{1}{2}\)

\(D=\left(\frac{1}{1.2}-\frac{1}{11.12}\right).\frac{1}{2}\)

\(D=\frac{65}{132}.\frac{1}{2}\)

\(D=\frac{65}{264}\)

19 tháng 5 2016

Giải:

Ta có nhận xét:

\(\frac{1}{1.2}-\frac{1}{2.3}=\frac{3-1}{1.2.3}=\frac{2}{1.2.3}\)

\(\frac{1}{2.3}-\frac{1}{3.4}=\frac{4-2}{2.3.4}=\frac{2}{2.3.4}\)

=>\(\frac{1}{1.2.3}=\frac{1}{3}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)\)

\(\frac{1}{2.3.4}=\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)\)

Do đó M=\(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{10.11}-\frac{1}{11.12}\right)\)

=\(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{11.12}\right)=\frac{1}{2}-\frac{1}{11.12}\)

=\(\frac{1}{2}.\frac{65}{132}=\frac{65}{124}\)

Vậy M=65/124

19 tháng 5 2016

M=\(\frac{65}{124}\)

20 tháng 3 2018

Ta có :

\(\dfrac{1}{1.2}-\dfrac{1}{2.3}=\dfrac{3}{1.2.3}-\dfrac{1}{1.2.3}=\dfrac{2}{1.2.3}\)

\(\dfrac{1}{2.3}-\dfrac{1}{3.4}=\dfrac{4}{2.3.4}-\dfrac{2}{2.3.4}=\dfrac{2}{2.3.4}\)

...

Do đó :

\(\dfrac{1}{1.2.3}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)\)

\(\dfrac{1}{2.3.4}=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)\)

Vậy :

\(M=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{11.12}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{132}\right)\)

\(=\dfrac{1}{2}.\dfrac{65}{132}=\dfrac{65}{264}\)

29 tháng 4 2020

      \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{10.11.12}\)

\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{1}{10.11.12}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{10.11}-\frac{1}{11.12}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{11.12}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{132}\right)\)

\(=\frac{1}{2}.\frac{65}{132}=\frac{65}{264}\)