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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm liên tục trên khoảng $\left( 0; +\infty \right)$, biết ${f}'\left( x \right)+\left( 2x+1 \right){{f}^{2}}\left( x \right)=0, {f}'\left( x \right)>0, \forall x>0$ và $f\left( 2 \right)=\dfrac{1}{6}$. Tính giá trị của $\left( P \right)=f\left( 1 \right)+f\left( x \right)+...+f\left( 2019 \right)$
A. $\dfrac{2019}{2020}$
B. $\dfrac{2018}{2019}$
C. $\dfrac{2021}{2020}$
D. $\dfrac{2020}{2019}$
A. $\dfrac{2019}{2020}$
B. $\dfrac{2018}{2019}$
C. $\dfrac{2021}{2020}$
D. $\dfrac{2020}{2019}$
TH1: $f\left( x \right)=0\Rightarrow {f}'\left( x \right)=0$ trái giả thiết
TH2: $f\left( x \right)\ne 0\Rightarrow {f}'\left( x \right)=-\left( 2x+1 \right).{{f}^{2}}\left( x \right)\Rightarrow \dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}=-\left( 2x+1 \right)$
$\Rightarrow \int{\dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}dx}=-\int{\left( 2x+1 \right)dx}\Rightarrow \dfrac{-1}{f\left( x \right)}=-\left( {{x}^{2}}+x+C \right)$
Ta có: $f\left( 2 \right)=\dfrac{1}{6}\Rightarrow C=0\Rightarrow f\left( x \right)=\dfrac{1}{{{x}^{2}}+x}=\dfrac{1}{x}-\dfrac{1}{x+1}$
$\Rightarrow P=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...-\dfrac{1}{2020}=\dfrac{2019}{2020}$
TH2: $f\left( x \right)\ne 0\Rightarrow {f}'\left( x \right)=-\left( 2x+1 \right).{{f}^{2}}\left( x \right)\Rightarrow \dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}=-\left( 2x+1 \right)$
$\Rightarrow \int{\dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}dx}=-\int{\left( 2x+1 \right)dx}\Rightarrow \dfrac{-1}{f\left( x \right)}=-\left( {{x}^{2}}+x+C \right)$
Ta có: $f\left( 2 \right)=\dfrac{1}{6}\Rightarrow C=0\Rightarrow f\left( x \right)=\dfrac{1}{{{x}^{2}}+x}=\dfrac{1}{x}-\dfrac{1}{x+1}$
$\Rightarrow P=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...-\dfrac{1}{2020}=\dfrac{2019}{2020}$
Đáp án A.