Cho 2 biểu thức sau A = 92 − 1/9 − 2/10 − 3/11 − . . . − 92/100 và B = 1/45 + 1/50 + 1/55 + . . . + 1/500 . Tính A/B .

\(A = 92 - \frac{1}{9} - \frac{2}{{10}} - \frac{3}{{11}} - ... - \frac{{92}}{{100}}\)

\( = \left( {1 - \frac{1}{9}} \right) + \left( {1 - \frac{2}{{10}}} \right) + \left( {1 - \frac{3}{{11}}} \right) + ... + \left( {1 - \frac{{92}}{{100}}} \right)\) (có 92 số hạng)

\( = \frac{8}{9} + \frac{8}{{10}} + \frac{8}{{11}} + ... + \frac{8}{{100}}\)

\( = 8.\left( {\frac{1}{9} + \frac{1}{{10}} + \frac{1}{{11}} + ... + \frac{1}{{100}}} \right)\)

\(\begin{array}{l}B = \frac{1}{{45}} + \frac{1}{{50}} + \frac{1}{{55}} + ... + \frac{1}{{500}}\\ = \frac{1}{{5.9}} + \frac{1}{{5.10}} + \frac{1}{{5.11}} + ... + \frac{1}{{5.100}}\\ = \frac{1}{5}.\frac{1}{9} + \frac{1}{5}.\frac{1}{{10}} + \frac{1}{5}.\frac{1}{{11}} + ... + \frac{1}{5}.\frac{1}{{100}}\\ = \frac{1}{5}.\left( {\frac{1}{9} + \frac{1}{{10}} + \frac{1}{{11}} + ... + \frac{1}{{100}}} \right)\end{array}\)

\(\frac{A}{B} = \frac{{8.\left( {\frac{1}{9} + \frac{1}{{10}} + \frac{1}{{11}} + ... + \frac{1}{{100}}} \right)}}{{\frac{1}{5}.\left( {\frac{1}{9} + \frac{1}{{10}} + \frac{1}{{11}} + ... + \frac{1}{{100}}} \right)}} = \frac{8}{{\frac{1}{5}}} = 8.5 = 40\)