Cho hàm số f(x) liên tục và dương trên

* Hướng dẫn giải

Đáp án A

Xét $f^{\text{'}}\left(x\right)+\left(2x+4\right)f^2\left(x\right)=0\iff \frac{-f^{\text{'}}\left(x\right)}{f^2\left(x\right)}=2x+4$

$\Rightarrow \int \frac{-f^{\text{'}}\left(x\right)}{f^2\left(x\right)}dx=\int \left(2x+4\right)dx\Rightarrow \frac{1}{f\left(x\right)}=x^2+4x+C$.

Vì $f\left(0\right)=\frac{1}{3}\Rightarrow C=3\Rightarrow f\left(x\right)=\frac{1}{x^2+4x+3}=\frac{1}{2}\left(\frac{1}{x+1}-\frac{1}{x+3}\right)$.

Vậy $S=\left[f\left(0\right)+f\left(2\right)+\mathrm{...}+f\left(2018\right)\right]+\left[f\left(1\right)+f\left(3\right)+\mathrm{...}+f\left(2017\right)\right]$

$$\begin{gathered} S=\frac{1}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\mathrm{...}+\frac{1}{2019}-\frac{1}{2021}\right]+ \\ \frac{1}{2}\left[\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\mathrm{...}+\frac{1}{2018}-\frac{1}{2020}\right] \end{gathered}$$

$S=\frac{1}{2}\left[1+\frac{1}{2}-\frac{1}{2020}-\frac{1}{2021}\right]=\frac{1}{2}\left[\frac{2020}{2021}+\frac{1009}{2020}\right]$

.