Ta có: $A = \frac{2}{{5.7}} + \frac{5}{{7.12}} + \frac{7}{{12.19}} + \frac{9}{{19.28}} + \frac{{11}}{{28.39}} + \frac{1}{{39.40}}$

$= \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{{12}} + \frac{1}{{12}} - \frac{1}{{19}} + \frac{1}{{19}} - \frac{1}{{28}} + \frac{1}{{28}} - \frac{1}{{39}} + \frac{1}{{39}} - \frac{1}{{40}}$

$= \frac{1}{5} - \frac{1}{{40}}$$= \frac{8}{{40}} - \frac{1}{{40}} = \frac{7}{{40}}$;

$B = \frac{1}{{20}} + \frac{1}{{44}} + \frac{1}{{77}} + \frac{1}{{119}} + \frac{1}{{170}}$

$= \frac{2}{{40}} + \frac{2}{{88}} + \frac{2}{{154}} + \frac{2}{{238}} + \frac{2}{{340}}$

$= 2.\left( {\frac{1}{{5.8}} + \frac{1}{{8.11}} + \frac{1}{{11.14}} + \frac{1}{{14.17}} + \frac{1}{{17.20}}} \right)$

$= 2.\frac{1}{3}.\left( {\frac{3}{{5.8}} + \frac{3}{{8.11}} + \frac{3}{{11.14}} + \frac{3}{{14.17}} + \frac{3}{{17.20}}} \right)$

$= \frac{2}{3}.\left( {\frac{1}{5} - \frac{1}{8} + \frac{1}{8} - \frac{1}{{11}} + \frac{1}{{11}} - \frac{1}{{14}} + \frac{1}{{14}} - \frac{1}{{17}} + \frac{1}{{17}} - \frac{1}{{20}}} \right)$

$= \frac{2}{3}.\left( {\frac{1}{5} - \frac{1}{{20}}} \right)$$= \frac{2}{3}.\left( {\frac{4}{{20}} - \frac{1}{{20}}} \right)$$= \frac{2}{3}.\frac{3}{{20}}$$= \frac{1}{{10}}$.

Ta có: $\frac{7}{{40}} > \frac{4}{{40}} = \frac{1}{{10}}$.

Do đó $A > B$.​