Cho hàm số \(f\left( x \right) = \ln \frac{{x + 1}}{{x + 4}}\). Tính giá trị biểu thức \(P = f'\left( 0 \right) + f'\left( 3 \right) + f'\left( 6 \right) + ... + f'\left( {2019} \r...

* Hướng dẫn giải

Ta có \(f(x)= \ln \dfrac{{x + 1}}{{x + 4}}\)\( \Rightarrow f'\left( x \right) = \dfrac{{{{\left( {\dfrac{{x + 1}}{{x + 4}}} \right)}'}}}{{\dfrac{{x + 1}}{{x + 4}}}} = \dfrac{\dfrac{3}{{{{\left( x + 4 \right)}^2}}}}{\dfrac{x + 1}{x + 4}}\)

\( \Rightarrow f'\left( x \right) = \dfrac{3}{{\left( {x + 1} \right)\left( {x + 4} \right)}}\)

\( \Rightarrow f'\left( x \right) = \dfrac{{\left( {x + 4} \right) - \left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x + 4} \right)}} = \dfrac{1}{{x + 1}} - \dfrac{1}{{x + 4}}\)

Khi đó \(P = f'\left( 0 \right) + f'\left( 3 \right) + f'\left( 6 \right) + ... + f'\left( {2019} \right)\)

\(\begin{array}{l} \Rightarrow P = 1 - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{7} + \dfrac{1}{7} - \dfrac{1}{{10}} + ... + \dfrac{1}{{2020}} - \dfrac{1}{{2023}}\\ \Leftrightarrow P = 1 - \dfrac{1}{{2023}} = \dfrac{{2022}}{{2023}}\end{array}\)

Chọn C.