Bạn tham khảo:
$\dfrac{1}{1.3} + \dfrac{1}{3.5} + \dfrac{1}{5.7} + ... + \dfrac{1}{x(x+2)} = \dfrac{2019}{4040}$
$⇔ 1 + ( - \dfrac{1}{3} + \dfrac{1}{3}) + (- \dfrac{1}{5} + \dfrac{1}{5} ) + ... + ( -\dfrac{1}{x} + \dfrac{1}{x} ) - \dfrac{1}{x+2} = \dfrac{2019}{4040}$
$⇔ 1 - \dfrac{1}{x + 2 } = \dfrac{2019}{4040}$
$⇔ x = 2018$