Cho hai biểu thức:

\[A = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{{2022}}\] và \[B = \frac{{2021}}{1} + \frac{{2020}}{2} + \frac{{2019}}{3} + ... + \frac{1}{{2021}}\].

Tính tỉ số \[\frac{B}{A}\].

a) \[\left( {\frac{{ - 1}}{6} + \frac{5}{{ - 12}}} \right) + \frac{7}{{12}}\]\[ = \frac{{ - 1}}{6} + \left( {\frac{5}{{ - 12}} + \frac{7}{{12}}} \right)\]

\[ = \frac{{ - 1}}{6} + \left( {\frac{{ - 5}}{{12}} + \frac{7}{{12}}} \right)\]\[ = \frac{{ - 1}}{6} + \frac{2}{{12}}\]\[ = \frac{{ - 1}}{6} + \frac{1}{6} = 0\];

b) \[\frac{7}{{36}} - \frac{8}{{ - 9}} + \frac{{ - 2}}{3}\]\[ = \frac{7}{{36}} + \frac{8}{9} + \frac{{ - 2}}{3}\]

\[ = \frac{7}{{36}} + \frac{{32}}{{36}} + \frac{{ - 24}}{{36}}\]\( = \frac{{15}}{{36}}\)\( = \frac{5}{{12}};\)

c) \[\frac{3}{5} - \frac{2}{5}.\frac{{10}}{{12}}\]\[ = \frac{3}{5} - \frac{1}{3}\]\[ = \frac{9}{{15}} - \frac{5}{{15}} = \frac{4}{{15}}\];

d) \[\frac{2}{{{{\left( { - 3} \right)}^2}}} + \frac{5}{{ - 12}} - \frac{{ - 3}}{4}\]\[ = \frac{2}{9} + \frac{{ - 5}}{{12}} + \frac{3}{4}\]

\[ = \frac{8}{{36}} + \frac{{ - 15}}{{36}} + \frac{{27}}{{36}}\]\[ = \frac{{20}}{{36}}\]\[ = \frac{5}{9}\].