\(1-\dfrac{2}{3.5}-\dfrac{2}{5.7}-\dfrac{2}{7.9}-...-\dfrac{2}{61.63}-\dfrac{2}{61.63}\)
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OY
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=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)
22 tháng 7 2021
Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}\)
\(=\dfrac{4}{15}\)
HN
3
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`
SI
21 tháng 6 2021
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{2015.2017}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)
\(=\dfrac{1}{3}-\dfrac{1}{2017}\)
\(=\dfrac{2017}{6051}-\dfrac{3}{6051}=\dfrac{2014}{6051}\)
DA
30 tháng 4 2022
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{2020.2022}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2020}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}\)
\(=\dfrac{2021}{2022}\)
23 tháng 7 2021
Ta có: \(1-\dfrac{2}{3\cdot5}-\dfrac{2}{5\cdot7}-\dfrac{2}{7\cdot9}-...-\dfrac{2}{61\cdot63}-\dfrac{2}{63\cdot65}\)
\(=1-\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{61\cdot63}+\dfrac{2}{63\cdot65}\right)\)
\(=1-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{61}-\dfrac{1}{63}+\dfrac{1}{63}-\dfrac{1}{65}\right)\)
\(=1-\left(\dfrac{1}{3}-\dfrac{1}{65}\right)\)
\(=1-\dfrac{62}{195}\)
\(=\dfrac{133}{195}\)
LS
2
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
\(B=\dfrac{34}{99}-\dfrac{1}{50}\)
\(B=\dfrac{1601}{4950}\)
TC
14 tháng 4 2022
\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=2\cdot\left(\dfrac{1}{3}-\dfrac{1}{101}\right)=2\cdot\dfrac{98}{303}=\dfrac{196}{303}\)
\(1-\dfrac{2}{3.5}-\dfrac{2}{5.7}-...-\dfrac{2}{61.63}-\dfrac{2}{63.65}\)
\(=1-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{63}-\dfrac{1}{65}\right)\)
\(=1-\left(\dfrac{1}{3}-\dfrac{1}{65}\right)\)
\(=1-\dfrac{62}{195}\)
\(=\dfrac{133}{195}\)
=1-(1/3-1/5+1/5-1/7+...+1/61-1/63)
=1-20/63=43/63