Tính \(A = \frac{1}{6} + \frac{1}{{12}} + \frac{1}{{20}} + \frac{1}{{30}} + \frac{1}{{42}} + \frac{1}{{56}} + \frac{1}{{72}} + \frac{1}{{90}}.\)

Ta có: \(A = \frac{1}{6} + \frac{1}{{12}} + \frac{1}{{20}} + \frac{1}{{30}} + \frac{1}{{42}} + \frac{1}{{56}} + \frac{1}{{72}} + \frac{1}{{90}}\)

\(A = \frac{1}{{2.3}} + \frac{1}{{3.4}} + \frac{1}{{4.5}} + \frac{1}{{5.6}} + \frac{1}{{6.7}} + \frac{1}{{7.8}} + \frac{1}{{8.9}} + \frac{1}{{9.10}}\)

\(A = \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \frac{1}{5} - \frac{1}{6} + \frac{1}{6} - \frac{1}{7} + \frac{1}{7} - \frac{1}{8} + \frac{1}{8} - \frac{1}{9} + \frac{1}{9} - \frac{1}{{10}}\)

\(A = \frac{1}{2} - \frac{1}{{10}}\)

\(A = \frac{2}{5}.\)

Vậy \(A = \frac{2}{5}.\)