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14 tháng 3 2017

\(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)

=\(\dfrac{1}{99.97}-\)(\(\dfrac{1}{97.95}+\dfrac{1}{95.93}+...+\dfrac{1}{5.3}+\dfrac{1}{3.1}\))

=\(\dfrac{1}{99.97}-\)\(\dfrac{1}{2}\left(\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{93}-\dfrac{1}{95}+\dfrac{1}{3}-\dfrac{1}{5}+1-\dfrac{1}{3}\right)\)

=\(\dfrac{1}{99.97}-\dfrac{1}{2}\left(1-\dfrac{1}{97}\right)\)

=\(\dfrac{1}{99.97}-\dfrac{1}{2}.\dfrac{96}{97}\)

=\(\dfrac{1}{99.97}-\dfrac{48}{97}\)

=\(\dfrac{1}{99.97}-\dfrac{48.99}{99.97}\)

=\(\dfrac{-4751}{9603}\)

19 tháng 3 2017

Đáp án là: -49/99

15 tháng 3 2017

a)\(\frac{1}{99.97}\)\(\frac{1}{97.95}\)\(\frac{1}{95.93}\)−…−\(\frac{1}{5.3}\)\(\frac{1}{3.1}\)

=\(\frac{1}{99.97}\)−(\(\frac{1}{97.95}\)+\(\frac{1}{95.93}\)+…+\(\frac{1}{5.3}\)+\(\frac{1}{3.1}\))

=\(\frac{1}{99.97}\)\(\frac{1}{2}\).(\(\frac{1}{95}\)\(\frac{1}{97}\)+\(\frac{1}{93}\)\(\frac{1}{95}\)+…+\(\frac{1}{3}\)\(\frac{1}{5}\)+1−\(\frac{1}{3}\))

=\(\frac{1}{99.97}\)\(\frac{1}{2}\).(1−\(\frac{1}{97}\))
=\(\frac{1}{99.97}\)\(\frac{1}{2}\).\(\frac{96}{97}\)

=\(\frac{1}{99.97}\)\(\frac{48}{97}\)

=\(\frac{1}{99.97}\)\(\frac{48.99}{99.97}\)

=\(\frac{-4751}{9603}\)

11 tháng 3 2017

\(T=\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-...-\dfrac{1}{5\cdot3}-\dfrac{1}{3\cdot1}\)

\(T=\dfrac{1}{99\cdot97}-\left(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{95\cdot97}\right)\)

Đặt \(A=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{95\cdot97}\)

\(A=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{95\cdot97}\right)\)

\(A=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)

\(A=\dfrac{1}{2}\left(1-\dfrac{1}{97}\right)=\dfrac{1}{2}\cdot\dfrac{96}{97}=\dfrac{48}{97}\)

Thay \(A\) vào \(T\) ta có:\(T=\dfrac{1}{99\cdot97}-\dfrac{48\cdot99}{97\cdot99}=\dfrac{-4751}{9603}\)

11 tháng 3 2017

Đặt \(A=\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)

\(A=\dfrac{1}{99.97}-\left(\dfrac{1}{97.95}+\dfrac{1}{95.93}+...+\dfrac{1}{5.3}+\dfrac{1}{3.1}\right)\)

\(A=\dfrac{1}{99.97}-\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{93.95}+\dfrac{1}{95.97}\right)\)

Đặt \(B=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{93.95}+\dfrac{1}{95.97}\)

\(2B=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{93.95}+\dfrac{2}{95.97}\)

\(2B=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{93}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{97}\)

\(2B=1-\dfrac{1}{97}\)

\(2B=\dfrac{96}{97}\)

\(B=\dfrac{96}{97}:2\)

\(B=\dfrac{48}{97}\)

\(\Rightarrow A=\dfrac{1}{99.97}-\dfrac{48}{97}\)

\(A=\dfrac{1}{99.97}-\dfrac{48.99}{97.99}\)

\(A=\dfrac{1-48.99}{99.97}\)

\(A=-\dfrac{4751}{9603}\)

Vậy \(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}=-\dfrac{4751}{9603}\)

20 tháng 9 2023

\(B=\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-\dfrac{1}{95\cdot93}-...-\dfrac{1}{3\cdot1}\)

\(B=-\left(\dfrac{1}{3\cdot1}+\dfrac{1}{5\cdot3}+...+\dfrac{1}{97\cdot99}\right)\)

\(2B=-\left(\dfrac{2}{3\cdot1}+\dfrac{2}{5\cdot3}+...+\dfrac{2}{99\cdot97}\right)\)

\(2B=-\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(2B=-\left(1-\dfrac{1}{99}\right)\)

\(2B=-\dfrac{98}{99}\)

\(B=-\dfrac{98}{198}\)

Cậu ơi, \(\dfrac{1}{99\cdot97}\) là dương mà sao lại đưa vào ngoặc âm tất cả vậy nhỉ?

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

18 tháng 10 2017

chết thiếu dấu =

28 tháng 3 2017

Đặt A=\(\dfrac{1}{99.97}-\dfrac{1}{97.95}-........-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)

=\(\dfrac{1}{99.97}-\left(\dfrac{1}{97.95}+\dfrac{1}{95.93}+......+\dfrac{1}{5.3}+\dfrac{1}{3.1}\right)\)

=\(\dfrac{1}{99.97}-\dfrac{1}{2}\left(\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{93}-\dfrac{1}{95}+.......+\dfrac{1}{3}-\dfrac{1}{5}+1-\dfrac{1}{3}\right)\) =\(\dfrac{1}{99.97}-\dfrac{1}{2}\left(1-\dfrac{1}{97}\right)\)

=\(\dfrac{1}{99.97}-\dfrac{1}{2}.\dfrac{96}{97}\)

=\(\dfrac{1}{99.97}-\dfrac{48}{97}\)

=\(\dfrac{1}{99.97}-\dfrac{48.99}{99.97}\)

=\(\dfrac{-4751}{9603}\)

28 tháng 3 2017

Đặt \(A=\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\\ \Rightarrow 2A= \dfrac{2}{99.97}-\dfrac{2}{97.95}-\dfrac{2}{95.93}-...-\dfrac{2}{5.3}-\dfrac{2}{3.1}\\ \Rightarrow 2A=\dfrac{1}{97}-\dfrac{1}{99}-(\dfrac{1}{95}-\dfrac{1}{97})-(\dfrac{1}{93}-\dfrac{1}{95})-...-(\dfrac{1}{1}-\dfrac{1}{3})\\ \Rightarrow 2A = \dfrac{1}{97}-\dfrac{1}{99}-(\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{93}-\dfrac{1}{95}+...+\dfrac{1}{1}-\dfrac{1}{3})\\ \Rightarrow 2A=\dfrac{1}{97}-\dfrac{1}{99}-1+\dfrac{1}{97}\\ \Rightarrow A\)