LN
2
K
Khách
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VD
14
31 tháng 1 2016
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+..........+\frac{1}{132}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..........+\frac{1}{11.12}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...........+\frac{1}{11}-\frac{1}{12}\)
\(=1-\frac{1}{12}\)
\(=\frac{11}{12}\)
5 tháng 7 2020
Bài làm
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}+\frac{1}{110}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(A=\frac{1}{1}-\frac{1}{11}\)
\(A=\frac{10}{11}\)
IK
3
3 tháng 8 2017
\(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{10.11}\)
=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-......+\frac{1}{10}-\frac{1}{11}.\)
=\(\frac{1}{2}-\frac{1}{11}\)
=\(\frac{9}{22}.\)
PN
2
25 tháng 5 2016
1/12+1/20+1/30+...+1/90+1/110
=1/3.4+1/4.5+1/5.6+...+1/9.10+1/10.11
=1/3-1/4+1/4-1/5+1/5-1/6+...+1/9-1/10+1/10-1/11
=1/3-1/11
=8/33
6 tháng 8 2016
1/12+1/20+1/30+...+1/90+1/110
=1/3.4+1/4.5+1/5.6+...+1/9.10+1/10.11
=1/3-1/4+1/4-1/5+1/5-1/6+...+1/9-1/10+1/10-1/11
=1/3-1/11
=8/33
NQ
3
DQ
2
25 tháng 7 2017
ta có:
1/6+1/12+1/20+1/30+.........+1/90+1/110
= 1/2x3+1/3x4+1/4x5+1/5x6+....+1/9x10+1/10x11
= 1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+....+1/9-1/10+1/10-1/11
=1/2-1/11=11/22-2/22=9/22
25 tháng 7 2017
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
\(=\left(\frac{1}{2}-\frac{1}{11}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...+\left(\frac{1}{10}-\frac{1}{10}\right)\)
\(=\frac{1}{2}-\frac{1}{11}=\frac{11}{22}-\frac{2}{22}=\frac{9}{22}\)
M
6
13 tháng 7 2017
misa
Đặt tên biểu thức là A ta có :
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+....+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{10}-\frac{1}{11}\)
\(=\left(\frac{1}{2}-\frac{1}{11}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+.....+\left(\frac{1}{10}-\frac{1}{10}\right)\)
\(=\left(\frac{1}{2}-\frac{1}{11}\right)+0+......+0\)
\(=\frac{11}{22}-\frac{2}{22}=\frac{9}{22}\)
13 tháng 7 2017
\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{10.11}\)
=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
=\(\frac{1}{2}-\frac{1}{11}\)
=\(\frac{9}{22}\)
14 tháng 5 2016
1/2+1/6+1/12+...+1/110
=1/1.2+1/2.3+1/3.4+...+1/10.11
=1-1/2+1/2-1/3+1/3-1/4+...+1/10-1/11
=1-1/11=10/11
ND
3
N
17 tháng 6 2017
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{9.10}+\frac{1}{10.11}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.......+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{2}-\frac{1}{11}\)
\(=\frac{9}{22}\)
17 tháng 6 2017
\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{10\cdot11}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}\cdot-\frac{1}{11}\)
\(=\frac{1}{2}-\frac{1}{11}\)
\(=\frac{9}{22}\)
\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{90}+\dfrac{1}{110}\)
\(=\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+...+\dfrac{1}{9x10}+\dfrac{1}{10x11}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\)
\(=\dfrac{1}{2}-\dfrac{1}{11}=\dfrac{11}{22}-\dfrac{2}{22}=\dfrac{9}{22}\)
1/6+1/12+1/20+1/90+1/110
=1/2x3+1/3x4+1/4x5+...+1/9x10+1/10x11
=1/2-1/3+1/3-1/4+1/4-1/5+1/5-...+1/9-1/10+1/10-1/11
=1/2-1/11=9/22