Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(M=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+.....+\frac{1}{10.11.12}\)
\(M=\frac{1}{2}-\frac{1}{11.12}\)
\(M=\frac{65}{132}\)
Ngắn gọn , xúc tích !!! :))
\(M=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\)
\(=\frac{1}{2}-\frac{1}{11.12}\)
\(=\frac{65}{132}\)
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
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 :
\(M=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{132}\right)\)
\(=\dfrac{1}{2}.\dfrac{65}{132}=\dfrac{65}{264}\)
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
M = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ... + 1/10.11.12
M = 1/2.(2/1.2.3 + 2/2.3.4 + 2/3.4.5 + ... + 2/10.11.12)
M = 1/2.(1/1.2 - 1/2.3 + 1/2.3- 1/3.4 + 1/3.4 - 1/4.5 + ... + 1/10.11 - 1/11.12)
M = 1/2.(1/1.2 - 1/11.12)
M = 1/4 - 1/264
M = 65/264
\(M=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+.....+\frac{1}{10.11}-\frac{1}{11.12}\)
\(M=\frac{1}{2}-\frac{1}{11.12}=\frac{65}{132}\)
\(M=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\right)\)
\(M=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{11.12}\right)\)
\(M=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{132}\right)\)
A = \(\frac{1}{1.2.3}\) + \(\frac{1}{2.3.4}\) + ...+ \(\frac{1}{10.11.12}\)
2 x A = \(\frac{2}{1.2.3}\) + \(\frac{2}{2.3.4}\) + ... + \(\frac{2}{10.11.12}\)
2A = \(\frac12\).(\(\frac{2}{1.3}\)) + \(\frac13\).(\(\frac{2}{2.4}\)) + ... + \(\frac{1}{11}\).(\(\frac{2}{10.12}\))
2A = \(\frac12\).(\(\frac11-\frac13\)) + \(\frac13\).(\(\frac12-\frac14\)) + ...+ \(\frac{1}{11}\).(\(\frac{1}{10}-\frac{1}{12}\))
2A = \(\frac{1}{1.2}\) - \(\frac{1}{2.3}\) + \(\frac{1}{2.3}\) - \(\frac{1}{3.4}\) + ...+\(\frac{1}{11.10}\) - \(\frac{1}{11.12}\)
2A = \(\frac{1}{1.2}-\frac{1}{11.12}\)
2A = \(\frac12-\frac{1}{132}\)
2A = \(\frac{66}{132}\) - \(\frac{1}{132}\)
2A = \(\frac{65}{132}\)
A = \(\frac{65}{132}\) : 2
A = \(\frac{65}{264}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{10.11.12}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{1}{10.11.12}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{10.11}-\frac{1}{11.12}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{11.12}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{132}\right)\)
\(=\frac{1}{2}.\frac{65}{132}=\frac{65}{264}\)
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
\(P=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{10.11.12}\)
\(P=\dfrac{1}{2}-\dfrac{1}{11.12}\)
\(P=\dfrac{65}{132}\)
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
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 \(M=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{1.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{11.12}\right)=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}.\dfrac{65}{132}=\dfrac{65}{264}\)
/n.(n+1)(n+2)=1/n*(1/(n+1)-1/(n+2)) = 1/n(n+1) - 1/n(n+2) = 1/n - 1/(n+1) - 1/2n + 1/2(n+2)
. . . ...... ... . . . = 1/2n - 1/(n+1) + 1/2(n+2)
như vậy ta có
1/1.2.3 = 1/2 - 1/2 + 1/6
1/2.3.4 = 1/4 - 1/3 + 1/8
1/3.4.5 = 1/6 - 1/4 + 1/10
1/4.5.6 = 1/8 - 1/5 + 1/12
.........................................
1/(n-1)n(n+1)= 1/2(n-1) - 1/n + 1/2(n+1)
1/n.(n+1)(n+2)= 1/2n - 1/(n+1) + 1/2(n+2)
=> tong = 1/4 - 1/2(n+1) + 1/2(n+2)
Hình như dòng thứ 4 của bạn có vấn đề chỗ 2( 10. 11. 12) đó có phải phân số hoh ???
Phan Tùng Lâm viết gì mà tùm lum lên zậy ???
!!!