Google
Câu hỏi: Cho hàm số $f(x)$ liên tục và dương trên $\left( 0;+\infty \right)$ thỏa mãn ${f}'\left( x \right)+\left( 2x+4 \right){{f}^{2}}\left( x \right)=0$ và $f\left( 0 \right)=\dfrac{1}{3}$. Tính tổng $S=f\left( 0 \right)+f\left( 1 \right)+f\left( 2 \right)+...+f\left( 2018 \right)=\dfrac{a}{b}$ với $a\in \mathbb{Z},b\in \mathbb{N},\dfrac{a}{b}$ tối giản. Khi đó $b-a=?$
A. $\dfrac{1}{2}\left( \dfrac{2020}{2021}+\dfrac{1009}{2020} \right)$.
B. $\dfrac{1}{2}\left( \dfrac{2020}{2021}-\dfrac{1009}{2020} \right)$.
C. $\dfrac{1}{2}\left( \dfrac{2020}{2021}+1 \right)$.
D. 2019.
A. $\dfrac{1}{2}\left( \dfrac{2020}{2021}+\dfrac{1009}{2020} \right)$.
B. $\dfrac{1}{2}\left( \dfrac{2020}{2021}-\dfrac{1009}{2020} \right)$.
C. $\dfrac{1}{2}\left( \dfrac{2020}{2021}+1 \right)$.
D. 2019.
Xét ${f}'\left( x \right)+\left( 2x+4 \right){{f}^{2}}\left( x \right)=0\Leftrightarrow \dfrac{-{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}=2x+4$
$\Rightarrow \int{\dfrac{-{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}dx=\int{\left( 2x+4 \right)dx\Rightarrow \dfrac{1}{f\left( x \right)}={{x}^{2}}+4x+C}}$.
Vì $f\left( 0 \right)=\dfrac{1}{3}\Rightarrow C=3\Rightarrow f\left( x \right)=\dfrac{1}{{{x}^{2}}+4x+3}=\dfrac{1}{2}\left( \dfrac{1}{x+1}-\dfrac{1}{x+3} \right)$.
Vậy $S=\left[ f\left( 0 \right)+f\left( 2 \right)+...+f\left( 2018 \right) \right]+\left[ f\left( 1 \right)+f\left( 3 \right)+...+f\left( 2017 \right) \right]$
$S=\dfrac{1}{2}\left[ 1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2019}-\dfrac{1}{2021} \right]+\dfrac{1}{2}\left[ \dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2018}-\dfrac{1}{2020} \right]$
$S=\dfrac{1}{2}\left[ 1+\dfrac{1}{2}-\dfrac{1}{2020}-\dfrac{1}{2021} \right]=\dfrac{1}{2}\left[ \dfrac{2020}{2021}+\dfrac{1009}{2020} \right]$.
$\Rightarrow \int{\dfrac{-{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}dx=\int{\left( 2x+4 \right)dx\Rightarrow \dfrac{1}{f\left( x \right)}={{x}^{2}}+4x+C}}$.
Vì $f\left( 0 \right)=\dfrac{1}{3}\Rightarrow C=3\Rightarrow f\left( x \right)=\dfrac{1}{{{x}^{2}}+4x+3}=\dfrac{1}{2}\left( \dfrac{1}{x+1}-\dfrac{1}{x+3} \right)$.
Vậy $S=\left[ f\left( 0 \right)+f\left( 2 \right)+...+f\left( 2018 \right) \right]+\left[ f\left( 1 \right)+f\left( 3 \right)+...+f\left( 2017 \right) \right]$
$S=\dfrac{1}{2}\left[ 1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2019}-\dfrac{1}{2021} \right]+\dfrac{1}{2}\left[ \dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2018}-\dfrac{1}{2020} \right]$
$S=\dfrac{1}{2}\left[ 1+\dfrac{1}{2}-\dfrac{1}{2020}-\dfrac{1}{2021} \right]=\dfrac{1}{2}\left[ \dfrac{2020}{2021}+\dfrac{1009}{2020} \right]$.
Đáp án A.