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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên khoảng $\left( 0;+\infty \right)$ và $f\left( x \right)\ne 0$ với mọi $x>0$. Tính tổng $f\left( 1 \right)+f\left( 2 \right)+...+f\left( 2022 \right)$ biết rằng ${f}'\left( x \right)=\left( 2x+1 \right){{f}^{2}}\left( x \right)$ và $f\left( 1 \right)=-\dfrac{1}{2}$
A. $\dfrac{2022}{2023}$.
B. $\dfrac{2021}{2022}$.
C. $-\dfrac{2021}{2022}$.
D. $\dfrac{-2022}{2023}$.
A. $\dfrac{2022}{2023}$.
B. $\dfrac{2021}{2022}$.
C. $-\dfrac{2021}{2022}$.
D. $\dfrac{-2022}{2023}$.
Ta có: $\dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}=2x+1$ $\Rightarrow -\dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}=-2x-1$ $\Rightarrow {{\left( \dfrac{1}{f\left( x \right)} \right)}^{\prime }}=-2x-1$
$\Rightarrow \dfrac{1}{f\left( x \right)}=-{{x}^{2}}-x+C$ $\Rightarrow \dfrac{1}{f\left( 1 \right)}=-2+C$ $\Rightarrow C=0$ $\Rightarrow f\left( x \right)=-\dfrac{1}{{{x}^{2}}+x}$ $=\dfrac{1}{x+1}-\dfrac{1}{x}$
Ta có: $f\left( 1 \right)+f\left( 2 \right)+...+f\left( 2022 \right)$
$=f\left( 2022 \right)+f\left( 2021 \right)+...+f\left( 1 \right)$ $=\left( \dfrac{1}{2023}-\dfrac{1}{2022} \right)+\left( \dfrac{1}{2022}-\dfrac{1}{2021} \right)+...+\left( \dfrac{1}{2}-1 \right)$
$=\dfrac{1}{2023}-1$ $=-\dfrac{2022}{2023}$.
$\Rightarrow \dfrac{1}{f\left( x \right)}=-{{x}^{2}}-x+C$ $\Rightarrow \dfrac{1}{f\left( 1 \right)}=-2+C$ $\Rightarrow C=0$ $\Rightarrow f\left( x \right)=-\dfrac{1}{{{x}^{2}}+x}$ $=\dfrac{1}{x+1}-\dfrac{1}{x}$
Ta có: $f\left( 1 \right)+f\left( 2 \right)+...+f\left( 2022 \right)$
$=f\left( 2022 \right)+f\left( 2021 \right)+...+f\left( 1 \right)$ $=\left( \dfrac{1}{2023}-\dfrac{1}{2022} \right)+\left( \dfrac{1}{2022}-\dfrac{1}{2021} \right)+...+\left( \dfrac{1}{2}-1 \right)$
$=\dfrac{1}{2023}-1$ $=-\dfrac{2022}{2023}$.
Đáp án D.