Giải:
\(\left(\dfrac{1}{2^2}-1\right).\left(\dfrac{1}{3^2}-1\right).....\left(\dfrac{1}{100^2}-1\right)\)
\(=\dfrac{-3}{2.2}.\dfrac{-8}{3.3}.....\dfrac{-9999}{100.100}\)
\(=\dfrac{1.-3.2.-4.....99.-101}{2.2.3.3.....100.100}\)
\(=\dfrac{\left(1.2.....99\right)}{\left(2.3.....100\right)}.\dfrac{\left(-3.-4.....-101\right)}{\left(2.3.....100\right)}\)
\(=\dfrac{1}{100}.\dfrac{-101}{2}\)
\(=\dfrac{-101}{200}\)
Chúc bạn học tốt!