Ta có: $\dfrac{1}{x}+\dfrac{1}{x.(x+1)}+\dfrac{1}{(x+1).(x+2)}+...+\dfrac{1}{(x+2018).(x+2019)}$
$=\dfrac{1}{x}+\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+2018}-\dfrac{1}{x+2019}$
$=\dfrac{2}{x}-\dfrac{1}{x+2019}$
$=\dfrac{4038+x}{x(x+2019}$