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10 tháng 9 2022

$= 2( \dfrac{1}{3.5} + \dfrac{5.7} + \dfrac{7.9} + ... + \dfrac{19.21})$
$= 2.\dfrac{1}{2}(\dfrac{1}{3} - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{7} + ... + \dfrac{1}{19} - \dfrac{1}{21})$

$= \dfrac{1}{3} - \dfrac{1}{21}$

$= \dfrac{7}{21} - \dfrac{1}{21}$
$= \dfrac{6}{21}$

$= \dfrac{2}{7}$

10 tháng 9 2022

$= 2( \dfrac{1}{3.5} + \dfrac{1}{5.7} + \dfrac{1}{7.9} + ... + \dfrac{1}{19.21})$
$= 2.\dfrac{1}{2}(\dfrac{1}{3} - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{7} + ... + \dfrac{1}{19} - \dfrac{1}{21})$

$= \dfrac{1}{3} - \dfrac{1}{21}$

$= \dfrac{7}{21} - \dfrac{1}{21}$
$= \dfrac{6}{21}$

$= \dfrac{2}{7}$

22 tháng 7 2021

Đặt A=\(\dfrac{2}{3.5}.\dfrac{2}{7.9}.....\dfrac{2}{99.101}\)

A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

A=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)

Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)

\(=\dfrac{1}{3}-\dfrac{1}{15}\)

\(=\dfrac{4}{15}\)

16 tháng 5 2017

\(M=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)

\(M=2.(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99})\)

\(M=2.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)

\(M=2.\dfrac{32}{99}\)

\(M=\dfrac{64}{99}\)

10 tháng 4 2018

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26 tháng 4 2022

\(B=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.100}\)

\(B=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99.100}+\dfrac{1}{99.100}\)

\(B=\dfrac{1}{3}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{100}\)

\(B=\dfrac{1}{3}-\dfrac{2}{100}+\dfrac{1}{99}\)

\(B=\dfrac{1}{3}-\dfrac{1}{50}+\dfrac{1}{99}\)

Đến đây thì hết tính hợp lý được rồi:v

\(B=\dfrac{34}{99}-\dfrac{1}{50}\)

\(B=\dfrac{1601}{4950}\)

 

26 tháng 4 2022

101 chứ bạn

21 tháng 6 2021

`2/(3.5)+2/(5.7)+....+2/(2015.2017)`

`=1/3-1/5+1/5-1/7+....+1/2016-1/2017`

`=1/3-1/2017=2014/6051`

21 tháng 6 2021

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{2015.2017}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)

\(=\dfrac{1}{3}-\dfrac{1}{2017}\)

\(=\dfrac{2017}{6051}-\dfrac{3}{6051}=\dfrac{2014}{6051}\)

30 tháng 4 2022

\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{2020.2022}\)

 

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2020}-\dfrac{1}{2022}\)

 

\(=1-\dfrac{1}{2022}\)

 

\(=\dfrac{2021}{2022}\)

30 tháng 4 2022

2/2*[2/1-2/2022]=2021/1011

15 tháng 6 2023

\(1-\dfrac{2}{3.5}-\dfrac{2}{5.7}-...-\dfrac{2}{61.63}-\dfrac{2}{63.65}\)

\(=1-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{63}-\dfrac{1}{65}\right)\)

\(=1-\left(\dfrac{1}{3}-\dfrac{1}{65}\right)\)

\(=1-\dfrac{62}{195}\)

\(=\dfrac{133}{195}\)

=1-(1/3-1/5+1/5-1/7+...+1/61-1/63)

=1-20/63=43/63

11 tháng 5 2022

ơi

11 tháng 5 2022

2 tháng 5 2017

\(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\)

\(A=\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+\dfrac{9-7}{7.9}+...+\dfrac{61-59}{59.61}\)

\(A=\dfrac{5}{3.5}-\dfrac{3}{3.5}+\dfrac{7}{5.7}-\dfrac{5}{5.7}+\dfrac{9}{7.9}-\dfrac{7}{7.9}+...+\dfrac{61}{59.61}-\dfrac{59}{59.61}\)

\(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\)

\(A=\dfrac{1}{3}-\dfrac{1}{61}=\dfrac{61}{183}-\dfrac{3}{183}=\dfrac{58}{183}\)

2 tháng 5 2017

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\)

= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\)

= \(\dfrac{1}{3}-\dfrac{1}{61}\)

= \(\dfrac{58}{183}\)

14 tháng 4 2022

\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{101}\right)=2\cdot\dfrac{98}{303}=\dfrac{196}{303}\)

14 tháng 4 2022

= 2/3 . 2/5 + 2/5 . 2/7 + ... + 2/99 . 2/101

= 2/3 - 2/5 + 2/5 - 2/7 + ... + 2/99 - 2/101

= 2/3 - 2/101

= 196/303

5 tháng 4 2017

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{39}\)

\(=(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13})+\dfrac{2}{39}\)

\(=(\dfrac{1}{3}-\dfrac{1}{13})+\dfrac{2}{39}\)

\(=\dfrac{10}{39}+\dfrac{2}{39}\)

\(=\dfrac{4}{13}\)

5 tháng 4 2017

gọi biểu thức đó là A

A=\(1.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{13}\right)+\dfrac{2}{39}\)

A= \(\left(\dfrac{1}{3}-\dfrac{1}{13}\right)+\dfrac{2}{39}=\dfrac{4}{13}\)

mk nhanh nhất nha bạn