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9 tháng 10 2021
\(=\dfrac{1}{99}-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{95\cdot97}+\dfrac{1}{97\cdot99}\right)\\ =\dfrac{1}{99}-\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{95\cdot97}+\dfrac{2}{97\cdot99}\right)\\ =\dfrac{1}{99}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\\ =\dfrac{1}{99}-\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)=\dfrac{1}{99}-\dfrac{1}{2}\cdot\dfrac{98}{99}\\ =\dfrac{1}{99}-\dfrac{49}{99}=-\dfrac{48}{99}=-\dfrac{16}{33}\)
CT
0
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 \(A=\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\\ \Rightarrow 2A= \dfrac{2}{99.97}-\dfrac{2}{97.95}-\dfrac{2}{95.93}-...-\dfrac{2}{5.3}-\dfrac{2}{3.1}\\ \Rightarrow 2A=\dfrac{1}{97}-\dfrac{1}{99}-(\dfrac{1}{95}-\dfrac{1}{97})-(\dfrac{1}{93}-\dfrac{1}{95})-...-(\dfrac{1}{1}-\dfrac{1}{3})\\ \Rightarrow 2A = \dfrac{1}{97}-\dfrac{1}{99}-(\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{93}-\dfrac{1}{95}+...+\dfrac{1}{1}-\dfrac{1}{3})\\ \Rightarrow 2A=\dfrac{1}{97}-\dfrac{1}{99}-1+\dfrac{1}{97}\\ \Rightarrow A\)
AN
2
4 tháng 4 2020
Ta có : \(\frac{1}{99.97}-\frac{1}{97.95}-...-\frac{1}{5.3}-\frac{1}{3.1}\)
= \(\frac{1}{99.97}-\left(\frac{1}{97.95}+...+\frac{1}{5.3}+\frac{1}{3.1}\right)\)
=\(\frac{1}{99.97}-\frac{1}{2}.\left(\frac{1}{95}-\frac{1}{97}+\frac{1}{93}-\frac{1}{95}+...+\frac{1}{3}-\frac{1}{5}+1-\frac{1}{3}\right)\)
= \(\frac{1}{99.97}-\frac{1}{2}.\left(1-\frac{1}{97}\right)\)
= \(\frac{1}{99.97}-\frac{1}{2}.\frac{96}{97}\)
= \(\frac{1}{99.97}-\frac{48}{97}=\frac{1}{99.97}-\frac{48.99}{99.97}=\frac{-4751}{9603}\)
4 tháng 4 2020
\(\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-...-\frac{1}{5.3}-\frac{1}{3.1}\)
\(=\frac{1}{99.97}-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{93.95}+\frac{1}{95.97}\right)\)
\(=\frac{1}{99.97}-\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{93.95}+\frac{2}{95.97}\right)\)
\(=\frac{1}{99.97}-\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{93}-\frac{1}{95}+\frac{1}{95}-\frac{1}{97}\right)\)
\(=\frac{1}{99.97}-\frac{1}{2}\left(1-\frac{1}{97}\right)\)
\(=\frac{1}{99.97}-\frac{1}{2}.\frac{96}{97}\)
\(=\frac{1}{9603}-\frac{48}{97}=\frac{-4751}{9603}\)
TZ
1
giúp mik nha!!!
\(=\dfrac{1}{99}-\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{97\cdot99}\right)\)
=1/99-1/2(1-1/3+1/3-1/5+...+1/97-1/99)
=1/99-1/2*98/99
=1/99-49/99
=-48/99=-16/33