\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{99^2}\right)\)
\(=\dfrac{2^2-1}{2^2}.\dfrac{3^2-1}{3^2}...\dfrac{99^2-1}{99^2}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}...\dfrac{98.100}{99.99}\)
\(=\dfrac{1}{2}.\dfrac{100}{99}=\dfrac{100}{198}\)
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