\(A=\left[\frac{1}{2^2}-1\right]\left[\frac{1}{3^2}-1\right]\left[\frac{1}{4^2}-1\right]\cdot...\cdot\left[\frac{1}{100^2}-1\right]\)
\(=\frac{-3}{2^2}\cdot\frac{-8}{3^2}\cdot\frac{-15}{4^2}\cdot...\cdot\frac{-9999}{100^2}\)
\(=\frac{-1\cdot3}{2\cdot2}\cdot\frac{-2\cdot4}{3\cdot3}\cdot\frac{-3\cdot5}{4\cdot4}\cdot...\cdot\frac{-99\cdot101}{100\cdot100}\)
\(=\frac{-1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot...\cdot100}\cdot\frac{3\cdot4\cdot5\cdot...\cdot101}{2\cdot3\cdot...\cdot100}\)
\(=-\frac{1}{100}\cdot\frac{101}{2}=-\frac{101}{200}\)
Mà \(-\frac{101}{200}< -\frac{1}{2}\)
nên \(A< -\frac{1}{2}\)
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{10000}-1\right)\)
\(A=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}...\frac{-9999}{10000}\)
\(A=\frac{-1.3}{2.2}.\frac{-2.4}{3.3}.\frac{-3.5}{4.4}...\frac{-99.101}{100.100}\)
\(A=\frac{\left(-1\right)\left(-2\right)\left(-3\right)...\left(-99\right)}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}\)
\(A=-\frac{1}{100}.\frac{101}{2}\)
\(A=-\frac{101}{200}\)
\(\text{Vậy A=}-\frac{101}{200}\)
