Giải bài 45 trang 41 SBT Toán 6 Cánh diều tập 2 - CD

Ta có:

\(\begin{array}{l}A = \frac{{{3^2}}}{{2.5}} + \frac{{{3^2}}}{{5.8}} + \frac{{{3^2}}}{{8.11}} = 3.\left( {\frac{3}{{2.5}} + \frac{3}{{5.8}} + \frac{3}{{8.11}}} \right) = 3.\left( {\frac{1}{2} - \frac{1}{5} + \frac{1}{5} - \frac{1}{8} + \frac{1}{8} - \frac{1}{{11}}} \right)\\ = 3.\left( {\frac{1}{2} - \frac{1}{{11}}} \right) = 3.\frac{9}{{22}} = \frac{{27}}{{22}} > 1\end{array}\)

\(\begin{array}{l}B = \frac{4}{{5.7}} + \frac{4}{{7.9}} + ... + \frac{4}{{59.61}} = 2.\left( {\frac{2}{{5.7}} + \frac{2}{{7.9}} + ... + \frac{2}{{59.61}}} \right)\\ = 2.\left( {\frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} + ... + \frac{1}{{59}} - \frac{1}{{61}}} \right) = 2.\left( {\frac{1}{5} - \frac{1}{{61}}} \right) = 2.\frac{{56}}{{305}} = \frac{{112}}{{305}} < 1\end{array}\)

 Vậy A >1 > B.