Ta có: \(B=\frac{1}{46}+\frac{2}{45}+\cdots+\frac{44}{3}+45\)
\(=\left(\frac{1}{46}+1\right)+\left(\frac{2}{45}+1\right)+\cdots+\left(\frac{44}{3}+1\right)+1\)
\(=\frac{47}{46}+\frac{47}{45}+\cdots+\frac{47}{3}+\frac{47}{47}=47\left(\frac13+\frac14+\cdots+\frac{1}{47}\right)\) =47A
=>\(\frac{A}{B}=\frac{1}{47}\)