Đáp án: đề thiếu , em xem lại nha, anh sửa lại đề đó
$\dfrac{1}{1.3}$ + $\dfrac{1}{3.5}$ + $\dfrac{1}{5.7}$ + $\dfrac{1}{7.9}$ + ... + $\dfrac{1}{2019.2021}$
= $\dfrac{1}{2}$ . [1- $\dfrac{1}{3}$ + $\dfrac{1}{3}$ - $\dfrac{1}{5}$ + $\dfrac{1}{5}$ - $\dfrac{1}{7}$ + $\dfrac{1}{7}$- $\dfrac{1}{9}$ + ... + (-$\dfrac{1}{2019}$) + $\dfrac{1}{2019}$ - $\dfrac{1}{2021}$ ]
= $\dfrac{1}{2}$ . [ 1 - $\dfrac{1}{2021}$]
= $\dfrac{1}{2}$ . [ $\dfrac{2021}{2021}$ -$\dfrac{1}{2021}$ ]
= $\dfrac{1}{2}$ . $\dfrac{2020}{2021}$
=$\dfrac{1010}{2021}$