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PN
13
3 tháng 1 2016
A=1/20+1/30+1/42+1/56+1/72+1/90
A=1/4x5+1/5x6+1/6x7+1/7x8+1/8x9+1/9x10
A=1/4-1/5+1/5-1/6+...+1/9-1/10
A=1/4-1/10
A=3/20
CT
2
13 tháng 9 2023
A=1/20+1/30+1/42+1/56+1/72+1/90
A=1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10
A=1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10
A=1/4-1/10
A=3/20
NK
0
LA
13 tháng 9 2018
A=1/20+1/30+1/42+1/56+1/72+1/90
A=1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10
A=1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10
A=1/4-1/10
A=3/20
k nha
13 tháng 9 2018
\(A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(A=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(A=\frac{1}{4}-\frac{1}{10}\)
\(A=\frac{3}{20}\)
Z
14 tháng 2 2018
1/72 + 1/56 + 1/42 + 1/30 + 1/20 + 1/12
= 1/( 8 .9 ) + 1/ ( 7 . 8 ) + 1/ ( 6 . 7 ) + 1/ ( 5 . 6 ) + 1/ ( 4 . 5 ) + 1/( 3 . 4 )
= 1/3 - 1/4 + 1/5 -1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9
= 1/3 - 1/9
= 3/9 - 1/9
= 2/9
Chắc 100%
14 tháng 2 2018
Ta có :
\(\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}\)
\(=\)\(\frac{1}{8.9}+\frac{1}{7.8}+\frac{1}{6.7}+\frac{1}{5.6}+\frac{1}{4.5}+\frac{1}{3.4}\)
\(=\)\(\frac{1}{8}-\frac{1}{9}+\frac{1}{7}-\frac{1}{8}+\frac{1}{6}-\frac{1}{7}+\frac{1}{5}-\frac{1}{6}+\frac{1}{4}-\frac{1}{5}+\frac{1}{3}-\frac{1}{4}\)
\(=\)\(-\frac{1}{9}+\frac{1}{3}\)
\(=\)\(\frac{-1}{9}+\frac{3}{9}\)
\(=\)\(\frac{2}{9}\)
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: \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{10\cdot11}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\)
\(=1-\dfrac{1}{11}=\dfrac{10}{11}\)
CC
6
HM
1
O
22 tháng 3 2022
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)
\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
15 tháng 5 2017
\(=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{11.12}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{2}-\frac{1}{12}\)
\(=\frac{5}{12}\)
15 tháng 5 2017
bn sẽ tinh theo kieeuranhaan 2 nha xin lỗi mik làm bi này rùi nhưng mik quên mik có sacks xem lại
`1/42 + 1/56 + 1/72 + .... + 1/9900`
`= 1/( 6*7) + 1/( 7*8 ) + ..... + 1/( 99*100)`
`= 1/6 - 1/7 + 1/7-1/8+....+1/99-1/100`
`= 1/6 - 1/100`
`= 50/300 - 3/300`
`= 47/300`
\(\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+...+\dfrac{1}{9900}\\ =\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+...+\dfrac{1}{99.100}\\ =\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{6}-\dfrac{1}{100}=\dfrac{47}{300}\)