Chia 2 vào 2 vế ta có
$\dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{20} + \cdots + \dfrac{1}{x(x+1)} = \dfrac{2019}{4042}$
Tách ra ta có
$\dfrac{1}{2.3} + \dfrac{1}{3.4} + \dfrac{1}{4.5} + \cdots + \dfrac{1}{x(x+1)} = \dfrac{2019}{4042}$
Tách ra
$\dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + \cdots + \dfrac{1}{x} - \dfrac{1}{x+1} = \dfrac{2019}{4042}$
Rút gọn
$\dfrac{1}{2} - \dfrac{1}{x+1} = \dfrac{2019}{4042}$
Chuyển vế ta có
$\dfrac{1}{x+1} = \dfrac{1}{2} - \dfrac{2019}{4042}$
Quy đồng
$\dfrac{1}{x+1} = \dfrac{1}{2021}$
Suy ra
$x + 1 = 2021$
Vậy
$x = 2020$
Nghiệm là $x = 2020$.
