A= 33.(\(\dfrac{1}{3}.\dfrac{3}{5}.\dfrac{5}{7}.....\dfrac{97}{99}\))
A=33.\(\left(\dfrac{1.3.5.....97}{3.5.7.....99}\right)=33.\dfrac{1}{99}=\dfrac{33.1}{99}=\dfrac{33}{99}=\dfrac{1}{3}\)
Vậy A = \(\dfrac{1}{3}\)
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