Đáp án+Giải thích các bước giải:
$A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2019}-\dfrac{1}{2020}+\dfrac{1}{2021}\\=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2019}+\dfrac{1}{2020}+\dfrac{1}{2021}\right)-\left(\dfrac22+\dfrac24+...+\dfrac 2{2020}\right)\\=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2019}+\dfrac{1}{2020}+\dfrac{1}{2021}\right)-\left(1+\dfrac12+....+\dfrac1{1010}\right)\\=B+\dfrac1{1010}$
$\left(A-B-1\right)^{2021}\\=(B+\dfrac1{1010}-B-1)^{2021}\\=\left(\dfrac{-1009}{1010}\right)^{2021}$
