a: \(A=\frac{4}{7\cdot31}+\frac{6}{7\cdot41}+\frac{9}{10\cdot41}+\frac{7}{10\cdot57}\)
\(=5\left(\frac{4}{31\cdot35}+\frac{6}{35\cdot41}+\frac{9}{41\cdot50}+\frac{7}{50\cdot57}\right)\)
\(=5\left(\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}\right)=5\left(\frac{1}{31}-\frac{1}{57}\right)\)
\(B=\frac{7}{19\cdot31}+\frac{5}{19\cdot43}+\frac{3}{23\cdot43}+\frac{11}{23\cdot57}\)
\(=2\left(\frac{7}{31\cdot38}+\frac{5}{38\cdot43}+\frac{3}{43\cdot46}+\frac{11}{46\cdot57}\right)\)
\(=2\left(\frac{1}{31}-\frac{1}{38}+\frac{1}{38}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}+\frac{1}{46}-\frac{1}{57}\right)\)
\(=2\left(\frac{1}{31}-\frac{1}{57}\right)\)
Do đó: \(\frac{A}{B}=\frac52\)
b: \(B=\frac{2020}{1}+\frac{2019}{2}+\frac{2018}{3}+\cdots+\frac{1}{2020}\)
\(=\left(1+\frac{2019}{2}\right)+\left(1+\frac{2018}{3}\right)+\cdots+\left(1+\frac{1}{2020}\right)+1\)
\(=\frac{2021}{2}+\frac{2021}{3}+\cdots+\frac{2021}{2021}=2021\left(\frac12+\frac13+\cdots+\frac{1}{2021}\right)=2021A\)
=>\(\frac{A}{B}=\frac{1}{2021}\)
