1/6+1/12+1/20+1/30+...+1/90+1/110
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5 tháng 7 2020
Bài làm
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}+\frac{1}{110}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(A=\frac{1}{1}-\frac{1}{11}\)
\(A=\frac{10}{11}\)
NT
2
15 tháng 7 2015
A= \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)
= \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{9x10}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
=\(1-\frac{1}{10}=\frac{9}{10}\)
15 tháng 7 2015
\(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}=\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{9\times10}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}=\frac{9}{10}\)
9 tháng 6 2016
1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2
= 1/90 - ( 1/72 + 1/56 + 1/42 + 1/30 + 1/20 + 1/12 + 1/6 + 1/2)
= 1/90 - ( 1/2 + 1/6 + 1/12 + ...+ 1/72)
= 1/90 - ( 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9)
= 1/90 - ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9)
= 1/90 - ( 1 - 1/9)
= 1/90 - 8/9
= 1/90 - 80/90
= -79/90
9 tháng 6 2016
1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2
= 1/90 - ( 1/72 + 1/56 + 1/42 + 1/30 + 1/20 + 1/12 + 1/6 + 1/2)
= 1/90 - ( 1/2 + 1/6 + 1/12 + ...+ 1/72)
= 1/90 - ( 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9)
= 1/90 - ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9)
= 1/90 - ( 1 - 1/9)
= 1/90 - 8/9
= 1/90 - 80/90
= -79/90
mk nha cac ban
IK
3
3 tháng 8 2017
\(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{10.11}\)
=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-......+\frac{1}{10}-\frac{1}{11}.\)
=\(\frac{1}{2}-\frac{1}{11}\)
=\(\frac{9}{22}.\)
25 tháng 8 2016
\(\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{1}{90}-\left(\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}-\frac{1}{72}\right)\)
\(=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\frac{8}{9}\)
\(=\frac{-79}{90}\)
25 tháng 8 2016
1/90 - 1/72 - 1/56 - ... - 1/6 - 1/2
= 1/90 - (1/2 + 1/6 + ... + 1/56 + 1/72)
= 1/90 - (1/1×2 + 1/2×3 + ... + 1/7×8 + 1/8×9)
= 1/90 - (1 - 1/2 + 1/2 - 1/3 + ... + 1/7 - 1/8 + 1/8 - 1/9)
= 1/90 - (1 - 1/9)
= 1/90 - 8/9
= 1/90 - 80/90
= -79/90
N
0
25 tháng 11 2015
1-1/2+1/2-1/3+1/3+1/4-1/4+1/5-1/5+1/6-1/6+1/7-1/7+1/8-1/8+1/9-1/9+1/10-(1-1/3+1/3-3/5+3/5-4/7+5/9-5/9+6/11-6/11-7/13)=1+1/10-1+7/13=83/130
TH
3 tháng 3 2020
\(D=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(D=\frac{1}{90}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}\right)\)
\(D=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)\)
\(D=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)
\(D=\frac{1}{90}-\frac{8}{9}=-\frac{79}{90}\)
3 tháng 3 2020
D=1/90 - 1/72 -1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2
D=1/90-(1/72+1/56+1/42+1/30+1/20+1/12+1/6+1/2)
D=1/90-(1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72)
D=1/90-(1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9)
D=1/90-(1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9)
D=1/90-(1/1-1/9)
D=1/90-8/9
D=(-79/90)
29 tháng 8 2023
a: \(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^7\)
=>\(2\cdot A=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^6\)
=>\(2A-A=1-\left(\dfrac{1}{2}\right)^7=1-\dfrac{1}{128}=\dfrac{127}{128}\)
=>\(A=\dfrac{127}{128}\)
b: \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{10\cdot11}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\)
\(=1-\dfrac{1}{11}=\dfrac{10}{11}\)
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
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
\(=\left(\frac{1}{2}-\frac{1}{11}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...+\left(\frac{1}{10}-\frac{1}{10}\right)\)
\(=\frac{1}{2}-\frac{1}{11}=\frac{11}{22}-\frac{2}{22}=\frac{9}{22}\)