Đáp án: $D$

Giải thích các bước giải:

Ta có: $e^{f(x)} = e^{-\ln (x^2 + x)}$ 

$= \dfrac{1}{e^{\ln (x^2 + x)}}$

$= \dfrac{1}{x^2 + x}$

$= \dfrac{1}{x(x + 1)}$

$= \dfrac{1}{x} - \dfrac{1}{x + 1}$

$\Rightarrow P = e^{f(1)} + e^{f(2)} + ... + e^{f(2019)}$
$= \dfrac{1}{1} - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + ... + \dfrac{1}{2019} - \dfrac{1}{2020}$

$= \dfrac{1}{1} + \bigg(\dfrac{1}{2} - \dfrac{1}{2}\bigg) + \bigg(\dfrac{1}{3} - \dfrac{1}{3}\bigg) + ... + \bigg(\dfrac{1}{2019} - \dfrac{1}{2019}\bigg) - \dfrac{1}{2020}$
$= 1 - \dfrac{1}{2020}$

$= \dfrac{2019}{2020}$

$\Rightarrow D$