Bn lm đc chx

chx ạ

Ta có: \(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2021\cdot2022}\)

\(=1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2021}-\frac{1}{2022}\)

\(=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{2022}-2\left(\frac12+\frac14+\cdots+\frac{1}{2022}\right)\)

\(=1+\frac12+\frac13+\cdots+\frac{1}{2022}-1-\frac12-\cdots-\frac{1}{1011}\)

\(=\frac{1}{1012}+\frac{1}{1013}+\cdots+\frac{1}{2022}\)

Ta có: \(B=1011+\frac{1010}{1012}+\frac{1009}{1013}+\cdots+\frac{2}{2020}+\frac{1}{2021}\)

\(=\left(\frac{1010}{1012}+1\right)+\left(\frac{1009}{1013}+1\right)+\cdots+\left(\frac{2}{2020}+1\right)+\left(\frac{1}{2021}+1\right)+1\)

\(=\frac{2022}{1012}+\frac{2022}{1013}+\cdots+\frac{2022}{2022}=2022\left(\frac{1}{1012}+\frac{1}{1013}+\cdots+\frac{1}{2022}\right)\)

=2022A

=>\(\frac{B}{A}=2022\) là số nguyên

Ta có: \(A-B\)

\(=1011\left(1+\frac13+\frac15+\cdots+\frac{1}{2019}\right)-1010\left(\frac12+\frac14+\cdots+\frac{1}{2020}\right)\)

\(=1010\left(1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2019}-\frac{1}{2020}\right)+\left(1+\frac13+\cdots+\frac{1}{2019}\right)\)

Vì \(1-\frac12>0;\frac13-\frac14>0;\ldots;\frac{1}{2019}-\frac{1}{2020}>0\)

nên \(1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2019}-\frac{1}{2020}>0\)

=>\(1010\left(1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2019}-\frac{1}{2020}\right)>0\)

=>A-B>0

=>A>B

\(\dfrac{2^{4042}\cdot3^{2020}}{6^{2019}\cdot4^{1011}}=\dfrac{2^{4042}\cdot3^{2020}}{3^{2019}\cdot2^{2019}\cdot2^{2022}}\)

\(=3\cdot\dfrac{2^{2042}}{2^{4041}}=3\cdot2=6\)