Google
Câu hỏi: Cho hàm số $f\left( x \right)\ne 0$ thỏa mãn điều kiện $f'\left( x \right)=\left( 2x+3 \right).{{f}^{2}}\left( x \right)$ và $f\left( 0 \right)=\dfrac{-1}{2}$. Biết tổng $f\left( 1 \right)+f\left( 2 \right)+...+f\left( 2017 \right)+f\left( 2018 \right)+f\left( 2019 \right)=\dfrac{a}{b}$ với $a\in \mathbb{Z},b\in \mathbb{N}*$ và $\dfrac{a}{b}$ là phân số tối giản. Mệnh đề nào sau đây đúng?
A. $\dfrac{a}{b}<-1.$
B. $\dfrac{a}{b}>1.$
C. $a+b=1010.$
D. $b-a=1516.$
A. $\dfrac{a}{b}<-1.$
B. $\dfrac{a}{b}>1.$
C. $a+b=1010.$
D. $b-a=1516.$
Biến đổi
$f'\left( x \right)=\left( 2x+3 \right).{{f}^{2}}\left( x \right)\Leftrightarrow \dfrac{f'\left( x \right)}{{{f}^{2}}\left( x \right)}=2x+3\Leftrightarrow \int{\dfrac{f'\left( x \right)}{{{f}^{2}}\left( x \right)}}dx=\int{\left( 2x+3 \right)}dx$
$\Leftrightarrow -\dfrac{1}{f\left( x \right)}={{x}^{2}}+3x+C\Rightarrow f\left( x \right)=-\dfrac{1}{{{x}^{2}}+3x+C}$. Mà $f\left( 0 \right)=-\dfrac{1}{2}$ nên $C=2$.
Do đó $f\left( x \right)=-\dfrac{1}{{{x}^{2}}+3x+2}=-\dfrac{1}{\left( x+1 \right)\left( x+2 \right)}.$
Khi đó $\dfrac{a}{b}=f\left( 1 \right)+f\left( 2 \right)+...+f\left( 2019 \right)+f\left( 2020 \right)$
$=-\left( \dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2020.2021}+\dfrac{1}{2021.2022} \right)$
$=-\left( \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022} \right)=-\left( \dfrac{1}{2}-\dfrac{1}{2022} \right)=\dfrac{-505}{1011}$
Với điều kiện a, b thỏa mãn bài toán, suy ra: $\left\{ \begin{aligned}
& a=-505 \\
& b=1011 \\
\end{aligned} \right.\Rightarrow b-a=1516.$
$f'\left( x \right)=\left( 2x+3 \right).{{f}^{2}}\left( x \right)\Leftrightarrow \dfrac{f'\left( x \right)}{{{f}^{2}}\left( x \right)}=2x+3\Leftrightarrow \int{\dfrac{f'\left( x \right)}{{{f}^{2}}\left( x \right)}}dx=\int{\left( 2x+3 \right)}dx$
$\Leftrightarrow -\dfrac{1}{f\left( x \right)}={{x}^{2}}+3x+C\Rightarrow f\left( x \right)=-\dfrac{1}{{{x}^{2}}+3x+C}$. Mà $f\left( 0 \right)=-\dfrac{1}{2}$ nên $C=2$.
Do đó $f\left( x \right)=-\dfrac{1}{{{x}^{2}}+3x+2}=-\dfrac{1}{\left( x+1 \right)\left( x+2 \right)}.$
Khi đó $\dfrac{a}{b}=f\left( 1 \right)+f\left( 2 \right)+...+f\left( 2019 \right)+f\left( 2020 \right)$
$=-\left( \dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2020.2021}+\dfrac{1}{2021.2022} \right)$
$=-\left( \dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022} \right)=-\left( \dfrac{1}{2}-\dfrac{1}{2022} \right)=\dfrac{-505}{1011}$
Với điều kiện a, b thỏa mãn bài toán, suy ra: $\left\{ \begin{aligned}
& a=-505 \\
& b=1011 \\
\end{aligned} \right.\Rightarrow b-a=1516.$
Đáp án D.