\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{1000}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{999}{1000}=\frac{1.2.3...999}{2.3.4...1000}=\frac{1}{1000}\)
\(B=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{2499}{2500}=\frac{3.8.15...2499}{4.9.16....2500}=\frac{1.3.2.4.3.5....49.51}{2.2.3.3.4.4...50.50}=\frac{\left(1.2.3...49\right).\left(3.4.5...51\right)}{\left(2.3.4...50\right).\left(2.3.4...50\right)}\)
\(\frac{1.51}{50.2}=\frac{51}{100}\)
a. \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{999}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{998}{999}\)
\(A=\frac{1\cdot2\cdot3\cdot....\cdot998}{2\cdot3\cdot4\cdot....\cdot999}=\frac{1}{999}\)
Vậy \(A=\frac{1}{999}\)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{1000}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{999}{1000}\)
\(=\frac{1.2.3.4....999}{2.3.4....1000}\)
\(B=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{2499}{2500}\)
\(=\frac{3.8.15....2499}{4.9.16....2500}\)
\(=\frac{1.3.2.4.3.5....49.51}{2.2.3.3.4.4....50.50}\)
\(=\frac{\left(1.2.3.4.5...49\right)\left(3.4.5....51\right)}{\left(2.3.4....50\right).\left(2.3.4...50\right)}\)
\(=\frac{1.51}{50.2}=\frac{51}{100}\)
\(=\frac{1}{1000}\)