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Câu hỏi: Cho hàm số ${f}'\left( x \right)=\left( 2x+1 \right).{{f}^{2}}\left( x \right)$ và $f\left( 1 \right)=-0,5$.
Tổng $f\left( 1 \right)+f\left( 2 \right)+f\left( 3 \right)+...+f\left( 2017 \right)+f\left( 2018 \right)+f\left( 2019 \right)+f\left( 2020 \right)=\dfrac{a}{b};\left( a\in \mathbb{Z};b\in \mathbb{N} \right)$ với $\dfrac{a}{b}$ tối giản. Khẳng định nào sau đây đúng?
A. $\dfrac{a}{b}<-1$.
B. $a\in \left( -2019;2019 \right)$.
C. $b-a=4041$.
D. $a+b=-1$.
Tổng $f\left( 1 \right)+f\left( 2 \right)+f\left( 3 \right)+...+f\left( 2017 \right)+f\left( 2018 \right)+f\left( 2019 \right)+f\left( 2020 \right)=\dfrac{a}{b};\left( a\in \mathbb{Z};b\in \mathbb{N} \right)$ với $\dfrac{a}{b}$ tối giản. Khẳng định nào sau đây đúng?
A. $\dfrac{a}{b}<-1$.
B. $a\in \left( -2019;2019 \right)$.
C. $b-a=4041$.
D. $a+b=-1$.
Ta có: ${f}'\left( x \right)=\left( 2x+1 \right).{{f}^{2}}\left( x \right)\Leftrightarrow \dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}=2x+1\Leftrightarrow \int{\dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}dx}=\int{\left( 2x+1 \right)dx}$
$\Leftrightarrow -\dfrac{1}{f\left( x \right)}={{x}^{2}}+x+C\Rightarrow \dfrac{1}{f\left( x \right)}=-{{x}^{2}}-x-C$.
Lại có: $f\left( 1 \right)=-0,5\Rightarrow -2=-{{1}^{2}}-1-C\Rightarrow C=0$.
Vậy $\dfrac{1}{f\left( x \right)}=-\left( {{x}^{2}}+x \right)=-x\left( x+1 \right)$ hay $-f\left( x \right)=\dfrac{1}{x\left( x+1 \right)}$.
Ta có: $-f\left( 1 \right)-f\left( 2 \right)-f\left( 3 \right)-...-f\left( 2017 \right)-f\left( 2018 \right)-f\left( 2019 \right)-f\left( 2020 \right)$
$=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{2018.2019}+\dfrac{1}{2019.2020}+\dfrac{1}{2020-2021}$
$=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2018}-\dfrac{1}{2019}+\dfrac{1}{2019}-\dfrac{1}{2020}+\dfrac{1}{2020}-\dfrac{1}{2021}=1-\dfrac{1}{2021}=\dfrac{2020}{2021}$.
Vậy $f\left( 1 \right)+f\left( 2 \right)+f\left( 3 \right)+...+f\left( 2020 \right)=\dfrac{-2020}{2021}$ hay $a=-2020,b=2021\Rightarrow b-a=4042$.
$\Leftrightarrow -\dfrac{1}{f\left( x \right)}={{x}^{2}}+x+C\Rightarrow \dfrac{1}{f\left( x \right)}=-{{x}^{2}}-x-C$.
Lại có: $f\left( 1 \right)=-0,5\Rightarrow -2=-{{1}^{2}}-1-C\Rightarrow C=0$.
Vậy $\dfrac{1}{f\left( x \right)}=-\left( {{x}^{2}}+x \right)=-x\left( x+1 \right)$ hay $-f\left( x \right)=\dfrac{1}{x\left( x+1 \right)}$.
Ta có: $-f\left( 1 \right)-f\left( 2 \right)-f\left( 3 \right)-...-f\left( 2017 \right)-f\left( 2018 \right)-f\left( 2019 \right)-f\left( 2020 \right)$
$=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{2018.2019}+\dfrac{1}{2019.2020}+\dfrac{1}{2020-2021}$
$=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2018}-\dfrac{1}{2019}+\dfrac{1}{2019}-\dfrac{1}{2020}+\dfrac{1}{2020}-\dfrac{1}{2021}=1-\dfrac{1}{2021}=\dfrac{2020}{2021}$.
Vậy $f\left( 1 \right)+f\left( 2 \right)+f\left( 3 \right)+...+f\left( 2020 \right)=\dfrac{-2020}{2021}$ hay $a=-2020,b=2021\Rightarrow b-a=4042$.
Đáp án C.