\(A=\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-....-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(A=\dfrac{1}{99}-\left(\dfrac{1}{99}-\dfrac{1}{98}\right)-\left(\dfrac{1}{98}-\dfrac{1}{97}\right)-....-\left(\dfrac{1}{3}-\dfrac{1}{2}\right)-\left(\dfrac{1}{2}-\dfrac{1}{1}\right)\)
(do \(\dfrac{n}{a\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với \(a\in N\)*)
\(A=\dfrac{1}{99}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{98}+\dfrac{1}{97}-......-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{2}+1\)
\(A=\dfrac{1}{99}-\dfrac{1}{99}+1=1\)
Vậy.........
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