1/99.97-1/97.95-.....-1/2.3
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HK
0
9 tháng 10 2021
\(=\dfrac{1}{99}-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{95\cdot97}+\dfrac{1}{97\cdot99}\right)\\ =\dfrac{1}{99}-\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{95\cdot97}+\dfrac{2}{97\cdot99}\right)\\ =\dfrac{1}{99}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\\ =\dfrac{1}{99}-\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)=\dfrac{1}{99}-\dfrac{1}{2}\cdot\dfrac{98}{99}\\ =\dfrac{1}{99}-\dfrac{49}{99}=-\dfrac{48}{99}=-\dfrac{16}{33}\)
18 tháng 10 2017
\(\frac{1}{99\cdot97}-\frac{1}{97\cdot95}-...-\frac{1}{5\cdot3}-\frac{1}{3\cdot1}\)\(=\frac{1}{99\cdot97}-\left(\frac{1}{97\cdot95}+\frac{1}{95\cdot93}+...+\frac{1}{3\cdot1}\right)\)
\(=\frac{1}{99\cdot97}-2\left(\frac{1}{97}-\frac{1}{95}+\frac{1}{95}-\frac{1}{93}+...+\frac{1}{3}-1\right)\)\(=\frac{1}{99\cdot97}-2\left(\frac{1}{97}-1\right)=\frac{1}{9603}-2\cdot\left(-\frac{96}{97}\right)\)\(\frac{1}{9603}-\frac{-192}{97}\)phần còn lại tự làm
28 tháng 3 2017
Đặt A=\(\dfrac{1}{99.97}-\dfrac{1}{97.95}-........-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
=\(\dfrac{1}{99.97}-\left(\dfrac{1}{97.95}+\dfrac{1}{95.93}+......+\dfrac{1}{5.3}+\dfrac{1}{3.1}\right)\)
=\(\dfrac{1}{99.97}-\dfrac{1}{2}\left(\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{93}-\dfrac{1}{95}+.......+\dfrac{1}{3}-\dfrac{1}{5}+1-\dfrac{1}{3}\right)\) =\(\dfrac{1}{99.97}-\dfrac{1}{2}\left(1-\dfrac{1}{97}\right)\)
=\(\dfrac{1}{99.97}-\dfrac{1}{2}.\dfrac{96}{97}\)
=\(\dfrac{1}{99.97}-\dfrac{48}{97}\)
=\(\dfrac{1}{99.97}-\dfrac{48.99}{99.97}\)
=\(\dfrac{-4751}{9603}\)
TV
28 tháng 3 2017
Đặt \(A=\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\\ \Rightarrow 2A= \dfrac{2}{99.97}-\dfrac{2}{97.95}-\dfrac{2}{95.93}-...-\dfrac{2}{5.3}-\dfrac{2}{3.1}\\ \Rightarrow 2A=\dfrac{1}{97}-\dfrac{1}{99}-(\dfrac{1}{95}-\dfrac{1}{97})-(\dfrac{1}{93}-\dfrac{1}{95})-...-(\dfrac{1}{1}-\dfrac{1}{3})\\ \Rightarrow 2A = \dfrac{1}{97}-\dfrac{1}{99}-(\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{93}-\dfrac{1}{95}+...+\dfrac{1}{1}-\dfrac{1}{3})\\ \Rightarrow 2A=\dfrac{1}{97}-\dfrac{1}{99}-1+\dfrac{1}{97}\\ \Rightarrow A\)
NN
2
H9
HT.Phong (9A5)
CTVHS
14 tháng 8 2023
Đặt: \(A=\dfrac{1}{99}-\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-....-\dfrac{1}{3\cdot1}\)
\(2A=\dfrac{2}{99}-\dfrac{2}{99\cdot97}-\dfrac{2}{97\cdot95}-...-\dfrac{2}{3\cdot1}\)
\(2A=\dfrac{2}{99}-\left(\dfrac{2}{99\cdot97}+\dfrac{2}{97\cdot95}+...+\dfrac{2}{3\cdot1}\right)\)
\(2A=\dfrac{2}{99}-\left(\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{95}-\dfrac{1}{97}+...+\dfrac{1}{3}-\dfrac{1}{5}+1-\dfrac{1}{3}\right)\)
\(2A=\dfrac{2}{99}-\left(-\dfrac{1}{99}+1\right)\)
\(2A=\dfrac{2}{99}-\dfrac{98}{99}\)
\(2A=-\dfrac{439}{99}\)
\(A=-\dfrac{439}{99}:2\)
\(A=-\dfrac{439}{198}\)
KV
14 tháng 8 2023
1/99 - 1/99.97 - 1/97.95 - ... - 1/3.1
= 1/99 - 1/2.(1/97 - 1/99 + 1/95 - 1/97 + ... + 1 - 1/3)
= 1/99 - 1/2.(1 - 1/99)
= 1/99 - 1/2 . 98/99
= 1/99 - 49/99
= -48/99
3 tháng 3 2022
\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{95\cdot97}\right)\)
\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)
\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\cdot\dfrac{96}{97}=\dfrac{-4751}{9603}\)
KK
0
30 tháng 9 2023
`#3107.101107`
\(B=\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-...-\dfrac{1}{5\cdot3}-\dfrac{1}{3\cdot1}\\ =\dfrac{1}{99\cdot97}-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{95\cdot97}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{97\cdot99}\right)-\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{95\cdot97}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{97}-\dfrac{1}{99}\right)-\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{97}-\dfrac{1}{99}\right)-\dfrac{1}{2}\cdot\left(1-\dfrac{1}{97}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{9603}-\dfrac{1}{2}\cdot\dfrac{96}{97}\\ =\dfrac{1}{2}\cdot\left(\dfrac{2}{9603}-\dfrac{96}{97}\right)\\ =\dfrac{1}{2}\cdot\left(-\dfrac{9502}{9603}\right)\\ =-\dfrac{4751}{9603}\)
Vậy, `B = -4751/9603.`
30 tháng 9 2023
\(B=\dfrac{1}{99.97}-\dfrac{1}{97.95}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
\(B=\dfrac{1}{97.99}-\left(\dfrac{1}{95.97}+...+\dfrac{1}{3.5}+\dfrac{1}{1.3}\right)\)
Đặt \(C=\dfrac{1}{95.97}+...+\dfrac{1}{3.5}+\dfrac{1}{1.3}\)
\(C=\dfrac{1}{95.97}+...+\dfrac{1}{3.5}+\dfrac{1}{1.3}\)
\(C=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{95.97}\)
\(C=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{95.97}\right):2\)
\(2C=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{95.97}\)
\(2C=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5} +...+\dfrac{1}{95}-\dfrac{1}{97}\)
\(2C=\dfrac{1}{1}-\dfrac{1}{97}\)
\(2C=\dfrac{96}{97}\)
\(C=\dfrac{96}{97}:2=\dfrac{48}{97}\)
Thay C vào ta được:
\(B=\dfrac{1}{97.99}-\dfrac{48}{97}\)
\(99B=\dfrac{99}{97.99}-\dfrac{48.99}{97}\)
\(99B=\dfrac{1}{97}-\dfrac{4752}{97}\)
\(99B=-\dfrac{4751}{97}\)
\(B=-\dfrac{4751}{97}:99=-\dfrac{4751}{9603}\)
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