Bạn tham khảo :
Giải thích :
Đặt $A=\dfrac{1}{99.97} - \dfrac{1}{97.95} - \dfrac{1}{95.93} - .... - \dfrac{1}{5.3} - \dfrac{1}{3.1}$
⇒ $2A =( \dfrac{2}{99.97} - \dfrac{2}{97.95} - \dfrac{2}{95.93} - .... - \dfrac{2}{5.3} - \dfrac{2}{3.1})$
⇒ $2A = 2( \dfrac{1}{1.3} + \dfrac{1}{3.5} + ... +\dfrac{1}{93.95} + \dfrac{1}{95.97} - \dfrac{1}{97.99})$
⇒ $2A = 2( 1 - \dfrac{1}{99})$
⇒ $2A = 2 . \dfrac{98}{99}$
⇒ $A = 2 . \dfrac{98}{99} : 2 = \dfrac{98}{99}$
