Cho hàm số f(x) khac 0, f'(x)=(2x+1) f^2 (x) và .f(1)=-0,5

* Hướng dẫn giải

Đáp án C

Ta có $f\text{'}\left(x\right)=\left(2x+1\right)f^2\left(x\right)\iff \frac{f\text{'}\left(x\right)}{f^2\left(x\right)}=2x+1\iff \int \frac{f\text{'}\left(x\right)}{f^2\left(x\right)}dx=\int \left(2x+1\right)dx$ 

$\iff \int \frac{d\left(f\left(x\right)\right)}{f^2\left(x\right)}=x^2+x+C\iff -\frac{1}{f\left(x\right)}=x^2+x+C\iff f\left(x\right)=-\frac{1}{x^2+x+C}$ 

Mà $f\left(1\right)=-\frac{1}{2}\Rightarrow -\frac{1}{C+2}=-\frac{1}{2}\iff C=0\to f\left(x\right)=-\frac{1}{x^2+x}=\frac{1}{x+1}-\frac{1}{x}$ 

$\Rightarrow f\left(1\right)+f\left(2\right)+\mathrm{...}+f\left(2017\right)=\frac{1}{2}-1+\frac{1}{3}-\frac{1}{2}+\mathrm{.....}+\frac{1}{2018}-\frac{1}{2017}=\frac{1}{2018}-1=\frac{a}{b}\Rightarrow \left\{\begin{array}{l}a=-2017\\ b=2018\end{array}\right.$ 

Vậy $b-a=2018-\left(-2017\right)=4035$