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B= \(\frac{1}{25.27}\)+ \(\frac{1}{27.29}\)+ \(\frac{1}{29.31}\)+ ..... + \(\frac{1}{73.75}\)
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DH
1
10 tháng 3 2016
\(\Rightarrow2A=\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\)
\(\Rightarrow2A=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\)
\(\Rightarrow2A=\frac{1}{25}-\frac{1}{75}=\frac{3}{75}-\frac{1}{75}=\frac{2}{75}\)
\(\Rightarrow A=\frac{2}{75}\div2=\frac{1}{75}\)
T
5 tháng 7 2019
#)Giải :
\(B=\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)
\(2B=\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\)
\(2B=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\)
\(2B=\frac{1}{25}-\frac{1}{75}\)
\(2B=\frac{2}{75}\)
\(B=\frac{2}{75}\div2\)
\(B=\frac{1}{75}\)
6 tháng 6 2015
A = \(\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
=\(7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
=\(7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
=\(7\left(\frac{1}{10}-\frac{1}{70}\right)\)
=\(7.\frac{3}{35}\)
=\(\frac{3}{5}\)
B=\(\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)
=\(\frac{1}{2}\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)
=\(\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)
=\(\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)\)
=\(\frac{1}{2}.\frac{2}{75}\)
=\(\frac{1}{75}\)
21 tháng 8 2021
`=>2B=(2)/(25.27)+(2)/(27.29)+(2)/(29.31)+....+(2)/(73.75)`
`=>2B=(1)/(25)-(1)/(27)+(1)/(27)-(1)/(29)+(1)/(29)-(1)/(31)+.....+(1)/(73)-(1)/(75)`
`=>2B=(1)/(25)-(1)/(75)`
`=>2B=(3)/(75)-(1)/(75)=(2)/(75)`
`=>B=(2)/(75):2`
`=>B=1/75`
21 tháng 8 2021
\(B=\dfrac{1}{25.27}+\dfrac{1}{27.29}+\dfrac{1}{29.31}+...+\dfrac{1}{73.75}\)
\(\Rightarrow2B=\dfrac{2}{25.27}+\dfrac{2}{27.29}+...+\dfrac{2}{73.75}=\dfrac{1}{25}-\dfrac{1}{27}+\dfrac{1}{27}-\dfrac{1}{29}+...+\dfrac{1}{73}-\dfrac{1}{75}\)\(\Rightarrow2B=\dfrac{1}{25}-\dfrac{1}{75}=\dfrac{2}{75}\Rightarrow B=\dfrac{1}{75}\)
21 tháng 1 2017
Giải:
a) C = \(\frac{6}{15.18}+\frac{6}{18.21}+...+\frac{6}{87.90}\)
C = \(\frac{6}{3}.\left(\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)
C = \(\frac{6}{3}.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
C = \(\frac{6}{3}.\left(\frac{1}{15}-\frac{1}{90}\right)\)
C = \(\frac{6}{3}.\frac{1}{18}\)
C = \(2.\frac{1}{18}\)
C = \(\frac{1}{9}\)
Vậy C = \(\frac{1}{9}\)
b) D = \(\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)
D = \(\frac{1}{2}.\left(\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\right)\)\
D = \(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)
D = \(\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)\)
D = \(\frac{1}{2}.\frac{2}{75}\)
D = \(\frac{1}{75}\)
Vậy D = \(\frac{1}{75}\)
c) E = \(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{38.41}\)
E = \(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{38}-\frac{1}{41}\)
E = \(\frac{1}{8}-\frac{1}{41}\)
E = \(\frac{33}{328}\)
Vậy E = \(\frac{33}{328}\)
TA
0
TT
18 tháng 5 2018
a,
suy ra A = 7. (1/10.11+1/11.12+1/12.13+.......+1/69.70)
suy ra A = 7. ( 1/10 - 1/11+ 1/11 - 1/12 + 1/12 - 1/13+ ............. + 1/69 - 1/70)
suy ra A = 7. ( 1/ 10 - 1/70)
suy ra A= 7. 3/35
suy ra A= 3/5
TK
3
10 tháng 7 2018
a, \(\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)
\(=\frac{1}{2}\left(\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\right)\)
\(=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)
\(=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)\)
\(=\frac{1}{2}\left(\frac{2}{75}\right)\)
\(=\frac{1}{75}\)
b, \(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=2\left(\frac{1004}{2010}\right)\)
\(=2\left(\frac{502}{1005}\right)\)
\(=\frac{1004}{1005}\)
Tk hộ =v
10 tháng 7 2018
\(\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}=\frac{1}{2}.\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)=\frac{1}{2}.\frac{2}{75}=\frac{1}{75}\)
\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)=2.\left(\frac{1}{2}-\frac{1}{2010}\right)=2.\frac{502}{1005}=\frac{1004}{1005}\)
B = \(\frac{1}{25}\)- \(\frac{1}{27}\)+ \(\frac{1}{27}\)-\(\frac{1}{29}\)+\(\frac{1}{29}\)-\(\frac{1}{31}\)+... + \(\frac{1}{73}\)-\(\frac{1}{75}\)=
B = \(\frac{1}{25}\)-\(\frac{1}{75}\)
B = \(\frac{2}{75}\)
Ủng hộ mik nha, mk đg âm điểm nè
B = 1/2.(1/25-1/27+1/27-1/29+....+1/73-1/75)
= 1/2.(1/25-1/75)
=1/2.2/75
= 1/75
kik cho mk nhé. đúng đấy. kb luôn