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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 ${f}'\left( x \right)+\left( 2x+3 \right).{{f}^{2}}\left( x \right)=0$, $f\left( x \right)>0$, $\forall x>0$ và $f\left( 1 \right)=\dfrac{1}{6}$. Tính giá trị của $P=1+f\left( 1 \right)+f\left( 2 \right)+f\left( 3 \right)+...+f\left( 2017 \right)$.
A. $\dfrac{6059}{4038}$.
B. $\dfrac{6055}{4038}$.
C. $\dfrac{6053}{4038}$.
D. $\dfrac{6047}{4038}$.
A. $\dfrac{6059}{4038}$.
B. $\dfrac{6055}{4038}$.
C. $\dfrac{6053}{4038}$.
D. $\dfrac{6047}{4038}$.
Giả thiết tương đương với: $\dfrac{-{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}=2x+3$.
Lấy nguyên hàm hai vế, ta được: $\int\limits_{{}}^{{}}{\dfrac{-{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}dx}=\int\limits_{{}}^{{}}{\left( 2x+3 \right)dx}$
$\Rightarrow \dfrac{1}{f\left( x \right)}={{x}^{2}}+3x+C\Rightarrow f\left( x \right)=\dfrac{1}{{{x}^{2}}+3x+C}\Rightarrow f\left( 1 \right)=\dfrac{1}{4+C}$
Mà $f\left( 1 \right)=\dfrac{1}{6}$, nên ta có $\dfrac{1}{4+C}=\dfrac{1}{6}\Rightarrow C=2\Rightarrow f\left( x \right)=\dfrac{1}{{{x}^{2}}+3x+2}=\dfrac{1}{x+1}-\dfrac{1}{x+2}$
$P=1+f\left( 1 \right)+f\left( 2 \right)+f\left( 3 \right)+...+f\left( 2017 \right)$
$=1+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2018}-\dfrac{1}{2019}=1+\dfrac{1}{2}-\dfrac{1}{2019}=\dfrac{6055}{4038}$.
Lấy nguyên hàm hai vế, ta được: $\int\limits_{{}}^{{}}{\dfrac{-{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}dx}=\int\limits_{{}}^{{}}{\left( 2x+3 \right)dx}$
$\Rightarrow \dfrac{1}{f\left( x \right)}={{x}^{2}}+3x+C\Rightarrow f\left( x \right)=\dfrac{1}{{{x}^{2}}+3x+C}\Rightarrow f\left( 1 \right)=\dfrac{1}{4+C}$
Mà $f\left( 1 \right)=\dfrac{1}{6}$, nên ta có $\dfrac{1}{4+C}=\dfrac{1}{6}\Rightarrow C=2\Rightarrow f\left( x \right)=\dfrac{1}{{{x}^{2}}+3x+2}=\dfrac{1}{x+1}-\dfrac{1}{x+2}$
$P=1+f\left( 1 \right)+f\left( 2 \right)+f\left( 3 \right)+...+f\left( 2017 \right)$
$=1+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2018}-\dfrac{1}{2019}=1+\dfrac{1}{2}-\dfrac{1}{2019}=\dfrac{6055}{4038}$.
Đáp án B.