\(\text{Ta có: }B=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+.....+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.....+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+....+\frac{1}{100}\left(1\right)\)
\(\text{Lại có:}A=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+.....+\frac{1}{100}\left(2\right)\)
\(\text{Từ (1) và (2) ta có A = B }\Rightarrow\frac{A}{B}=1\)
