Ta có A=\(\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)....\left(\frac{1}{400}-1\right)\)
=\(\frac{-3}{2^2}.\frac{-8}{3^2}.\frac{-15}{4^2}...\frac{-399}{20^2}\)
=\(\frac{-\left(1.3\right)}{2.2}.\frac{-\left(2.4\right)}{3.3}.\frac{-\left(3.5\right)}{4.4}....\frac{-\left(19.21\right)}{20.20}\)
=\(-\left(\frac{1.2.3...19}{2.3.4...20}.\frac{3.4.5...21}{2.3.4...20}\right)\)
=\(-\left(\frac{1}{20}.\frac{21}{2}\right)=-\frac{21}{40}< -\frac{21}{42}=-\frac{1}{2}\)