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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$, $f\left( x \right)\ne 0{}_{{}}\forall x\in \mathbb{R},{f}'\left( x \right)=\left( 2x+1 \right){{f}^{2}}\left( x \right)$ và $f\left( 1 \right)=-\dfrac{1}{2}$. Biết rằng tổng $f\left( 1 \right)+f\left( 2 \right)+f\left( 3 \right)+...+f\left( 2017 \right)=\dfrac{a}{b};\left( a\in \mathbb{Z},b\in \mathbb{N} \right)$ với $\dfrac{a}{b}$ tối giản. Mệnh đề nào dưới đây đúng?
A. $a+b=-1$
B. $a\in \left( -2017;2017 \right)$
C. $\dfrac{a}{b}<-1$
D. $b-a=4035$
A. $a+b=-1$
B. $a\in \left( -2017;2017 \right)$
C. $\dfrac{a}{b}<-1$
D. $b-a=4035$
Ta có ${f}'\left( x \right)=\left( 2x+1 \right){{f}^{2}}\left( x \right)\Leftrightarrow \dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}dx=\left( 2x+1 \right)\Rightarrow \int{\dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}dx=}\int{\left( 2x+1 \right)dx}$
$\Leftrightarrow -\dfrac{1}{f\left( x \right)}={{x}^{2}}+x+C$
Mà $f\left( 1 \right)=-\dfrac{1}{2}$ nên $C=0\Rightarrow f\left( x \right)=-\dfrac{1}{{{x}^{2}}+x}=\dfrac{1}{x+1}-\dfrac{1}{x}$. Mặt khác:
$\begin{aligned}
& f\left( 1 \right)+f\left( 2 \right)+f\left( 3 \right)+...+f\left( 2017 \right)=\left( \dfrac{1}{2}-1 \right)+\left( \dfrac{1}{3}-\dfrac{1}{2} \right)+\left( \dfrac{1}{4}-\dfrac{1}{3} \right)+...+\left( \dfrac{1}{2018}-\dfrac{1}{2017} \right) \\
& \Leftrightarrow f\left( 1 \right)+f\left( 2 \right)+f\left( 3 \right)+...+f\left( 2017 \right)=-1+\dfrac{1}{2018}=\dfrac{-2017}{2018} \\
\end{aligned}$
$\Rightarrow a=-2017;b=2018$. Khi đó $b-a=4035$
$\Leftrightarrow -\dfrac{1}{f\left( x \right)}={{x}^{2}}+x+C$
Mà $f\left( 1 \right)=-\dfrac{1}{2}$ nên $C=0\Rightarrow f\left( x \right)=-\dfrac{1}{{{x}^{2}}+x}=\dfrac{1}{x+1}-\dfrac{1}{x}$. Mặt khác:
$\begin{aligned}
& f\left( 1 \right)+f\left( 2 \right)+f\left( 3 \right)+...+f\left( 2017 \right)=\left( \dfrac{1}{2}-1 \right)+\left( \dfrac{1}{3}-\dfrac{1}{2} \right)+\left( \dfrac{1}{4}-\dfrac{1}{3} \right)+...+\left( \dfrac{1}{2018}-\dfrac{1}{2017} \right) \\
& \Leftrightarrow f\left( 1 \right)+f\left( 2 \right)+f\left( 3 \right)+...+f\left( 2017 \right)=-1+\dfrac{1}{2018}=\dfrac{-2017}{2018} \\
\end{aligned}$
$\Rightarrow a=-2017;b=2018$. Khi đó $b-a=4035$
Đáp án D.