Đáp án đúng: D
Phương pháp giải:
-Tính đạo hàm \(f'\left( x \right)\) theo công thức \(\left( {\ln u} \right)' = \dfrac{{u'}}{u}\).
- Thay \(x = 2,3,...,2019\) để tính \(f'\left( 2 \right) + f'\left( 3 \right) + ... + f'\left( {2019} \right)\).Giải chi tiết:Với \(x > 0\) ta có:
\(\begin{array}{l}f\left( x \right) = \ln \left( {\dfrac{{{x^2} - 1}}{{{x^2}}}} \right) = \ln \left( {x - 1} \right) + \ln \left( {x + 1} \right) - 2\ln x\\ \Rightarrow f'\left( x \right) = \dfrac{1}{{x - 1}} + \dfrac{1}{{x + 1}} - \dfrac{2}{x}\\ \Rightarrow f'\left( 2 \right) = \dfrac{1}{1} + \dfrac{1}{3} - \dfrac{2}{2}\\\,\,\,\,\,\,\,f'\left( 3 \right) = \dfrac{1}{2} + \dfrac{1}{4} - \dfrac{2}{3}\\\,\,\,\,\,\,f'\left( 4 \right) = \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{2}{4}\\\,\,\,\,\,\,...\\\,\,\,\,\,f'\left( {2019} \right) = \dfrac{1}{{2018}} + \dfrac{1}{{2020}} - \dfrac{2}{{2019}}\\ \Rightarrow f'\left( 2 \right) + f'\left( 3 \right) + ... + f'\left( {2019} \right)\\ = \left( {1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{{2018}}} \right) + \left( {\dfrac{1}{3} + \dfrac{1}{4} + ... + \dfrac{1}{{2020}}} \right) - 2\left( {\dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{{2019}}} \right)\\ = \left( {1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{{2018}}} \right) - \left( {\dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{{2019}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left( {\dfrac{1}{3} + \dfrac{1}{4} + ... + \dfrac{1}{{2020}}} \right) - \left( {\dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{{2019}}} \right)\\ = 1 - \dfrac{1}{{2019}} - \dfrac{1}{2} + \dfrac{1}{{2020}} = \dfrac{{2039190 - 1}}{{4078380}}\\ \Rightarrow m = 2039190,\,\,n = 4078380\end{array}\)
Chọn D.