$\text{Giải thích các bước giải:}$
$\text{Đặt}$ $A = \dfrac{1}{1×3} + \dfrac{1}{3×5} + \dfrac{1}{5×7} + ... + \dfrac{1}{2003×2005}$
$⇒ 2A = 2 × (\dfrac{1}{1×3} + \dfrac{1}{3×5} + \dfrac{1}{5×7} + ... + \dfrac{1}{2003×2005})$
$⇒ 2A = \dfrac{2}{1×3} + \dfrac{2}{3×5} + \dfrac{2}{5×7} + ... + \dfrac{2}{2003×2005}$
$⇒ 2A = 1 - \dfrac{1}{3} + \dfrac{1}{1×3} - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{7} + ... + \dfrac{1}{2003} - \dfrac{1}{2005}$
$⇒ 2A = 1 - \dfrac{1}{2005}$
$⇒ 2A = \dfrac{2005}{2005} - \dfrac{1}{2005}$
$⇒ 2A = \dfrac{2004}{2005}$
$⇒ A = \dfrac{1002}{2005}$
$\text{Chúc bạn học tốt !}$
