Đáp án:$\frac{{2019}}{{2020}}$
Giải thích các bước giải: $\eqalign{
& \ln \left( {\frac{{2018x}}{{x + 1}}} \right) = \ln (2018x) - \ln (x + 1) \cr
& \ln '\left( {\frac{{2018x}}{{x + 1}}} \right) = \ln '(2018x) - \ln '(x + 1) = \frac{1}{x} - \frac{1}{{x + 1}} \cr
& f'(1) = 1 - \frac{1}{2} \cr
& f'(2) = \frac{1}{2} - \frac{1}{3} \cr
& f'(3) = \frac{1}{3} - \frac{1}{4} \cr
& ... \cr
& f'(2019) = \frac{1}{{2019}} - \frac{1}{{2020}} \cr
& S = f'(1) + f'(2) + f'(3) + ... + f'(2019) = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{{2019}} - \frac{1}{{2020}} = 1 - \frac{1}{{2020}} = \frac{{2019}}{{2020}} \cr} $
