Ta có B =\(\frac{1}{51.100}+\frac{1}{52.99}+...+\frac{1}{100.51}\)
=> 151B = \(\frac{151}{51.100}+\frac{151}{52.99}+...+\frac{151}{100.51}=\frac{1}{51}+\frac{1}{100}+\frac{1}{52}+\frac{1}{99}+...+\frac{1}{51}+\frac{1}{100}\)
\(=2\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\)
=> B = \(\frac{2}{151}.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)\)
Khi đó \(\frac{A}{B}=\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{\frac{2}{151}.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\right)}=\frac{1}{\frac{2}{151}}=\frac{151}{2}=75,5\)
