Cho hàm số f'(x)=(2x1)f^2(x) vàf(1)=-0,5 . Tổng

Đáp án C

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

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

Lại có: $f\left(1\right)=-\mathrm{0,5}\Rightarrow -2=-1^2-1-C\Rightarrow C=0$.

Vậy $\frac{1}{f\left(x\right)}=-\left(x^2+x\right)=-x\left(x+1\right)$ hay $-f\left(x\right)=\frac{1}{x\left(x+1\right)}$.

Ta có: $-f\left(1\right)-f\left(2\right)-f\left(3\right)-\mathrm{...}-f\left(2017\right)-f\left(2018\right)-f\left(2019\right)-f\left(2020\right)$

          $=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{2018.2019}+\frac{1}{2019.2020}+\frac{1}{2020-2021}$

          $=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\mathrm{...}+\frac{1}{2018}-\frac{1}{2019}+\frac{1}{2019}-\frac{1}{2020}+\frac{1}{2020}-\frac{1}{2021}=1-\frac{1}{2021}=\frac{2020}{2021}$.

Vậy $f\left(1\right)+f\left(2\right)+f\left(3\right)+\mathrm{...}+f\left(2020\right)=\frac{-2020}{2021}$ hay $a=-\mathrm{2020,}b=2021\Rightarrow b-a=4042$.