Ta có A < \(\frac{2}{3^2-1^2}+\frac{2}{5^2-1^2}+...+\frac{2}{2019^2-1^2}\)
Tới đây ở mẫu số ta có công thức :
a2 - b2 = a2 - ab + ab - b2 = a(a - b) + b(a - b) = (a + b)(a - b)
<=> \(A< \frac{2}{\left(3-1\right)\left(3+1\right)}+\frac{2}{\left(5-1\right)\left(5+1\right)}+....+\frac{2}{\left(2019-1\right)\left(2019+1\right)}\)
\(=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2018.2020}=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2018}-\frac{1}{2020}\)
\(=\frac{1}{2}-\frac{1}{2020}=\frac{1009}{2020}< \frac{2019}{2020}=B\)
=> A < B
