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K
Khách
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30 tháng 3 2019
a) \(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)
\(=4.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\right)\)
\(=4.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=4.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=4.\frac{32}{99}\)
\(=\frac{128}{99}\)
KN
30 tháng 3 2019
\(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)
\(=2\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=2.\frac{32}{99}\)
\(=\frac{64}{99}\)
YS
4
27 tháng 1 2016
mk làm đc,vs điều kiện bạn phải tick cho mk khi mk giải xog, ko đc tick cho ai
25 tháng 4 2018
Ta có :
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(A=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)
\(A=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)
\(A=\frac{3}{2}.\frac{50}{51}\)
\(A=\frac{25}{17}\)
Vậy \(A=\frac{25}{17}\)
Chúc bạn học tốt ~
25 tháng 4 2018
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(A=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)
\(A=\frac{3}{2}.\frac{50}{51}\)
\(A=\frac{25}{17}\)
\(B=\frac{21}{4}\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(B=\frac{21}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(B=\frac{21}{4}\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)\)
\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(B=\frac{21}{4}.33.\frac{4}{21}\)
\(B=\left(\frac{21}{4}.\frac{4}{21}\right).33\)
\(B=33\)
\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(C=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(C=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(C=\frac{1}{2}.\frac{98}{99}\)
\(C=\frac{49}{99}\)
25 tháng 4 2017
ta co : 65%=0,65
goi A= 4.(1/3.5+1/5.7+1/7.9+............+1/97.99)
2A=4.( 2/3.5+2/5.7+2/7.9+...............+2/97.99)
2A=4.(1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99)
2A=4.(1/3-1/99)
2A=4.(33/=99+1/99)
2A=4.34/99
2A=136/99
A=136/99:2
A=68/99=0,69=0,68
Vi A=0,68 > 0,65
=> A > 65%
31 tháng 3 2017
M = \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{2015.2017}\)4/1.3 + 4/3.5 + 4/5.7 + ... + 4/2015.2017
M = \(2.\frac{2}{1.3}+2.\frac{2}{3.5}+2.\frac{2}{5.7}+...+2.\frac{2}{2015.2017}\) 2 . 2/1.3 + 2 . 2/3.5 + 2 . 2/5.7 + ... + 2 . 2/2015.2017
M = 2 . ( 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/2015.2017 )
M = 2 . ( 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/2015 - 1/2017 )
M = 2 . ( 1 - 1/2017 )
M = 2 . 2016/2017
M = 4032/2017
31 tháng 3 2017
\(M=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(M=2\left(1-\frac{1}{2017}\right)\)
\(M=\frac{4032}{2017}\)
TV
1
20 tháng 2 2017
de y \(\frac{1}{3}\)-\(\frac{1}{5}\)=\(\frac{2}{3.5}\)
tuong tu suy ra
M=\(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+......+\(\frac{1}{97}\)-\(\frac{1}{99}\)
M=\(\frac{1}{3}\)-\(\frac{1}{99}\)
M=\(\frac{32}{99}\)
D
6
EC
8 tháng 8 2016
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{3}.\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{3}.\frac{98}{99}\)
\(=\frac{98}{297}\)
Chuc bn học tốt
8 tháng 8 2016
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{99}\)
\(=1-\frac{1}{99}\)
\(=\frac{98}{99}\)
Ta có :
\(\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{97.99}\)
\(=\)\(2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\right)\)
\(=\)\(2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\)\(2\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=\)\(2.\frac{32}{99}\)
\(=\)\(\frac{64}{99}\)
Chúc bạn học tốt ~