-Bài này cũng là lớp 8 hả?
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2021^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2020.2021}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2020}-\dfrac{1}{2021}=1-\dfrac{1}{2021}=\dfrac{2020}{2021}< 1\)\(\Rightarrowđpcm\)