f(x)=Ln 2018x/x+1 tính S=f'(1) +f'(2)+…+f'(2019) 10/08/2021 Bởi Kylie f(x)=Ln 2018x/x+1 tính S=f'(1) +f'(2)+…+f'(2019)
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} $ Bình luận
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} $