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`A=1/(1.3)+1/(3.5)+....+1/(2009.2011)`
`=> 2A=2/(1.3)+1/(3.5)+...+2/(2009.2011)`
`=1-1/3+1/3+1/5+.....+1/2009-1/2011`
`=1-1/2011`
`=2010/2011`
`=> A=1005/2011`
Dựa trên công thức: \(\dfrac{a}{n.\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a}\), ta có:
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2009.2011}\)
= \(\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2009.2011}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)
=\(\dfrac{1}{2}.\left(1-\dfrac{1}{2011}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{2010}{2011}\right)\)
= \(\dfrac{2010}{4022}\)
= tự rút gọn nhé
Hok tốt!
\(S=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{99.100.101}\)
\(S=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{99.100}-\dfrac{1}{100.101}\right)\)
\(S=\dfrac{1}{4}-\dfrac{1}{2.100.101}\)
Đặt :
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...........+\dfrac{1}{2010.2011}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...........+\dfrac{1}{2010}-\dfrac{1}{2011}\)
\(A=1-\dfrac{1}{2011}\)
\(A=\dfrac{2010}{2011}\)
~ Chúc bn học tốt ~
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2010.2011}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2010}-\dfrac{1}{2011}\)
\(=\dfrac{1}{1}-\dfrac{1}{2011}\)
\(=\dfrac{2010}{2011}\)
Ta có: \(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{199\cdot200}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{199}-\dfrac{1}{200}\)
\(=\dfrac{1}{2}-\dfrac{1}{200}\)
\(=\dfrac{100-1}{200}=\dfrac{99}{200}< \dfrac{100}{200}=\dfrac{1}{2}\)(đpcm)
Ta có: \(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{199\cdot200}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\cdot\cdot\cdot+\dfrac{1}{199}-\dfrac{1}{200}\)
\(=\dfrac{1}{2}-\dfrac{1}{200}< \dfrac{1}{2}\) (Đpcm)
Biến đổi thừa số tổng quát: \(1+\dfrac{1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{\left(k-1\right)\left(k+1\right)+1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{k^2}{\left(k-1\right)\left(k+1\right)}\).
Do đó \(1+\dfrac{1}{1.3}=\dfrac{2^2}{1.3}\), \(1+\dfrac{1}{2.4}=\dfrac{3^2}{2.4}\), \(1+\dfrac{1}{3.5}=\dfrac{4^2}{3.5}\),..., \(1+\dfrac{1}{2018.2020}=\dfrac{2019^2}{2018.2020}\), \(1+\dfrac{1}{2019.2021}=\dfrac{2020^2}{2019.2021}\). Từ đó suy ra \(\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)...\left(1+\dfrac{1}{2019.2021}\right)\)
\(=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}.\dfrac{6^2}{5.7}...\dfrac{2019^2}{2018.2020}.\dfrac{2020^2}{2019.2021}\)
\(=\dfrac{2.2020}{2021}=\dfrac{4040}{2021}\)
2/1*2*3+2/3*4*5+...+2/2009*2010*2011
A=2/2*(1/1-1/2-1/3+1/2-1/3-1/4+1/4-1/5-1/6+...+1/2009-1/2010-1/2011
A=1*(1-1/2011)
A=1*2010/2011=2010/2011
suy ra: 2010/2011<1
suy ra 1/2 của 1 lớn hơn 2010/2011
VẬY A NHỎ HƠN 1/2
VẬY
