\(A=\left(1+\dfrac{1999}{1}\right)\left(1+\dfrac{1999}{2}\right)...\left(1+\dfrac{1999}{1000}\right)\)
\(=\dfrac{2000}{1}.\dfrac{2001}{2}.\dfrac{2002}{3}...\dfrac{2999}{1000}\)\(=\dfrac{2000.2001.2002...2999}{1.2.3...1000}\)
\(B=\left(1+\dfrac{1000}{1}\right)\left(1+\dfrac{1000}{2}\right)...\left(1+\dfrac{1000}{1999}\right)\)
\(=\dfrac{1001}{1}.\dfrac{1002}{2}.\dfrac{1003}{3}...\dfrac{2999}{1999}\) \(=\dfrac{1001.1002.1003...2999}{1.2.3...1999}\)
\(\Rightarrow A:B=\left(\dfrac{2000.2001.2002...2999}{1.2.3...1000}\right):\left(\dfrac{1001.1002.1003...2999}{1.2.3...1999}\right)\)
\(=\dfrac{2000.2001.2002...2999}{1.2.3...1000}.\dfrac{1.2.3...1999}{1001.1002.1003...2999}\)
\(=\dfrac{2000.2001.2002...2999}{1.2.3...1000}.\dfrac{1.2.3...1000.\left(1001.1002...1999\right)}{1001.1002.1003....1999.\left(2000.2001.2002.2999\right)}\)\(=\dfrac{1.2.3...1000}{1.2.3...1000}=1\)
Vậy \(\dfrac{A}{B}=1\)
