Rút gọn biểu thức sau:
S = \(\frac{1}{1.2.3}\)+\(\frac{1}{2.3.4}\)+\(\frac{1}{3.4.5}\)+ ... + \(\frac{1}{998.999.1000}\)
K
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18 tháng 4 2019
\(C=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{998\cdot999\cdot1000}\)
\(C=\frac{1}{2}\left[\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{998\cdot999\cdot1000}\right]\)
\(C=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{998\cdot999}-\frac{1}{999\cdot1000}\right]\)
\(C=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{999\cdot1000}\right]\)
Tính nốt :v
18 tháng 4 2019
Ta có
\(C=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{998\cdot999\cdot1000}\)
\(\Rightarrow2C=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{998\cdot999\cdot1000}\)
\(=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{998\cdot999}-\frac{1}{999\cdot1000}\)
\(=\frac{1}{1\cdot2}-\frac{1}{999\cdot1000}\)
\(=\frac{1}{2}-\frac{1}{999000}\)
\(=\frac{499500}{999000}-\frac{1}{999000}\)
\(=\frac{499499}{999000}\)
\(\Rightarrow C=\frac{499499}{1998000}\)
đúng nha bạn nhớ k mik
Q
14 tháng 3 2016
\(M=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\)
\(=\frac{1}{2}-\frac{1}{11.12}\)
\(=\frac{65}{132}\)
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
18 tháng 2 2016
2A=2(1/1.2.3+1/2.3.4+...+1/98.99.100)
2A=1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-...+1/98.99-1/99.100
2A=1/1.2-1/99.100
2A=4949/9900
A=4949/9900:2
A=4949/19800
Vậy A=4949/198000
TT
4 tháng 3 2020
Ta có : \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)
\(\Leftrightarrow2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{18.19.20}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{18.19}-\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{19.20}=\frac{189}{380}\)
\(\Rightarrow B=\frac{189}{760}\)
4 tháng 3 2020
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{380}\right)\)
\(=\frac{1}{2}.\frac{189}{380}=\frac{189}{760}\)
23 tháng 3 2017
Ta có A = \(\frac{1.2.3-2.3.4+3.4.5-4.5.6+5.6.7-6.7.8}{2.4.6-4.6.8+6.8.10-8.10.12+10.12.14-12.14.16}\)
A = \(\frac{1.2.3-2.3.4+3.4.5-4.5.6+5.6.7-6.7.8}{\left(1.2.3\right).2-\left(2.3.4\right).2+\left(3.4.5\right).2-\left(4.5.6\right).2+\left(5.6.7\right).2-\left(6.7.8\right).2}\)
A = \(\frac{1.\left(1.2.3-2.3.4+3.4.5-4.5.6+5.6.7-6.7.8\right)}{2.\left(1.2.3-2.3.4+3.4.5-4.5.6+5.6.7-6.7.8\right)}\)
A = \(\frac{1}{2}\)
VA
1
11 tháng 4 2019
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(=\frac{1}{2}.\frac{4949}{9900}\)
\(=\frac{4949}{19800}\)
PH
3
15 tháng 5 2019
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{2018\cdot2019\cdot2020}\)
\(=\frac{1}{2}\left[\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{2018\cdot2019\cdot2020}\right]\)
\(=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}-\frac{1}{2019\cdot2020}\right]\)
Đến đây tự tính được rồi:v
BD
15 tháng 5 2019
Đặt tổng trên là A
Ta có:
\(2A=2\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{2018\cdot2019\cdot2020}\right)\)
\(=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{2018\cdot2019\cdot2020}\)
\(=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}-\frac{1}{2019\cdot2020}\)
\(=\frac{1}{2}-\frac{1}{2019\cdot2020}\)
\(A=\left(\frac{1}{2}-\frac{1}{2019\cdot2020}\right)\div2\)
*Làm tiếp*
\(#Louis\)
PT
4
14 tháng 4 2019
A=1/2 *(1/1*2-1/2*3+1/2*3-1/3*4+........+1/98*99-1/99*100)
=1/2*(1/2-1/99*100)
=1/2*(4950-1/9900)
=4950/19800
14 tháng 4 2019
\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\)
\(A=\frac{1}{2}\left[\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{98\cdot99\cdot100}\right]\)
\(A=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+....+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right]\)
\(A=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{99\cdot100}\right]=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)
Giải :
\(\text{S}=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{998\cdot999\cdot1000}\)
\(\text{S}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{998}-\frac{1}{999}+\frac{1}{999}-\frac{1}{1000}\)
\(\text{S}=1-\frac{1}{1000}=\frac{999}{1000}\)
\(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{998.999.1000}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{998.999.1000}\right)\)
\(=\frac{1}{2}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{1000-998}{998.999.1000}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{998.999}-\frac{1}{999.1000}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{999.1000}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{999000}\right)\)
\(=\frac{1}{2}.\frac{499499}{999000}\)
\(=\frac{499499}{1998000}\)
Study well ! >_<