Đặt A=\(\dfrac{1}{99.97}-\dfrac{1}{97.95}-........-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
=\(\dfrac{1}{99.97}-\left(\dfrac{1}{97.95}+\dfrac{1}{95.93}+......+\dfrac{1}{5.3}+\dfrac{1}{3.1}\right)\)
=\(\dfrac{1}{99.97}-\dfrac{1}{2}\left(\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{93}-\dfrac{1}{95}+.......+\dfrac{1}{3}-\dfrac{1}{5}+1-\dfrac{1}{3}\right)\) =\(\dfrac{1}{99.97}-\dfrac{1}{2}\left(1-\dfrac{1}{97}\right)\)
=\(\dfrac{1}{99.97}-\dfrac{1}{2}.\dfrac{96}{97}\)
=\(\dfrac{1}{99.97}-\dfrac{48}{97}\)
=\(\dfrac{1}{99.97}-\dfrac{48.99}{99.97}\)
=\(\dfrac{-4751}{9603}\)
Đặt \(A=\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\\ \Rightarrow 2A= \dfrac{2}{99.97}-\dfrac{2}{97.95}-\dfrac{2}{95.93}-...-\dfrac{2}{5.3}-\dfrac{2}{3.1}\\ \Rightarrow 2A=\dfrac{1}{97}-\dfrac{1}{99}-(\dfrac{1}{95}-\dfrac{1}{97})-(\dfrac{1}{93}-\dfrac{1}{95})-...-(\dfrac{1}{1}-\dfrac{1}{3})\\ \Rightarrow 2A = \dfrac{1}{97}-\dfrac{1}{99}-(\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{93}-\dfrac{1}{95}+...+\dfrac{1}{1}-\dfrac{1}{3})\\ \Rightarrow 2A=\dfrac{1}{97}-\dfrac{1}{99}-1+\dfrac{1}{97}\\ \Rightarrow A\)
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