K
Khách
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KM
4
25 tháng 8 2015
A=-(1/1.3+1/3.5+1/5.7+...+1/97.99)
A=-1/2.(2/1.3+2/3.5+2/5.7+...+2/97.99)
A=-1/2.(1-1/3+1/3-1/5+...+1/97-1/99)
A=-1/2.(1-1/99)=-1/2.98/99
A=(tự bấm máy tính nha)
KK
0
NV
1
MH
14 tháng 4 2016
=> -A = \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{95.97}-\frac{1}{97.99}\)
=> -2A = \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{95.97}-\frac{2}{97.99}\)
=> \(-2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{95}-\frac{1}{97}-\frac{1}{97}+\frac{1}{99}\)
=> \(-2A=1-\frac{1}{97}-\frac{1}{97}+\frac{1}{99}=\frac{9502}{9603}\)
=> \(A=\frac{9502}{9603}:\left(-2\right)=-\frac{4751}{9603}\)
HN
1
NB
30 tháng 8 2019
\(\frac{1}{99}-\frac{1}{99.97}-\frac{1}{97.95}-...-\frac{1}{5.3}-\frac{1}{3}\\ =\frac{1}{99}-\left(\frac{1}{99.97}+\frac{1}{97.95}+...+\frac{1}{5.3}+\frac{1}{3.1}\right)\\ =\frac{1}{99}-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{95.97}+\frac{1}{97.99}\right)\\ =\frac{1}{99}-\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{95}-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\right)\\ =\frac{1}{99}-\frac{1}{2}.\left(1-\frac{1}{99}\right)\\ =\frac{1}{99}-\frac{1}{2}.\frac{98}{99}\\ =\frac{-16}{33}\)
16 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
\(A=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\)
`=`\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
`=`\(\dfrac{1}{3}-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-...-\dfrac{1}{9}\)
`=`\(\dfrac{1}{3}-\dfrac{1}{9}\)
`=`\(\dfrac{2}{9}\)
Vậy, \(A=\dfrac{2}{9}\)
`b)`
\(B=\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{23\cdot24}+\dfrac{1}{24\cdot25}\)
`=`\(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
`=`\(\dfrac{1}{5}-\left(\dfrac{1}{6}-\dfrac{1}{6}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\dfrac{1}{25}\)
`=`\(\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)
Vậy, \(B=\dfrac{4}{25}\)
`c)`
\(C=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)
`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
`=`\(1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-...-\dfrac{1}{100}\)
`=`\(1-\dfrac{1}{100}=\dfrac{99}{100}\)
Vậy, \(C=\dfrac{99}{100}\)
29 tháng 3 2023
a: \(A=\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{2022\cdot2024}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2022}-\dfrac{1}{2024}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{1011}{2024}=\dfrac{1011}{4848}< \dfrac{1}{4}\)
b: \(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2013\cdot2015}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2013}-\dfrac{1}{2015}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2014}{2015}=\dfrac{1007}{2015}< \dfrac{1}{2}\)
Ta có \(A=\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-...-\dfrac{1}{3\cdot1}\)
\(\Leftrightarrow2A=\dfrac{2}{99\cdot97}-\dfrac{2}{97\cdot95}-...-\dfrac{2}{3\cdot1}\)
\(=-\dfrac{1}{99}+\dfrac{1}{97}-\dfrac{1}{97}+\dfrac{1}{95}-...-\dfrac{1}{3}+1\)
\(=-\dfrac{1}{99}+1=\dfrac{98}{99}\)
\(\Rightarrow A=\dfrac{49}{99}\)
\(A=\dfrac{1}{99.97}-\dfrac{1}{97.95}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
\(=\dfrac{1}{99.97}-\left(\dfrac{1}{97.95}+...+\dfrac{1}{5.3}+\dfrac{1}{3.1}\right)\)
Đặt \(B=\dfrac{1}{97.95}+...+\dfrac{1}{5.3}+\dfrac{1}{3.1}\)
\(B=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{95.97}\)
\(2B=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{95.97}\)
\(2B=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{95}-\dfrac{1}{97}\)
\(2B=1-\dfrac{1}{97}\)
\(2B=\dfrac{96}{97}\)
\(B=\dfrac{96}{97}:2=\dfrac{48}{97}\)
\(\Rightarrow A=\dfrac{1}{99.97}-B=\dfrac{1}{9603}-\dfrac{48}{97}=\dfrac{-4751}{9603}\)