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Câu hỏi: Cho hàm số $f\left(x \right)=\ln \left(1-\dfrac{1}{{{x}^{2}}} \right)$. Biết rằng $f'\left(2 \right)+f'\left(3 \right)+...+f'\left(2019 \right)+f'\left(2020 \right)=\dfrac{m}{n}$ với $m$, $ n$ là các số nguyên dương nguyên tố cùng nhau. Tính $S=2m-n$.
A. $2$.
B. $4$.
C. $-2$.
D. $-4$.
A. $2$.
B. $4$.
C. $-2$.
D. $-4$.
Điều kiện. $1-\dfrac{1}{{{x}^{2}}}>0\Leftrightarrow \left\{ \begin{aligned}
& {{x}^{2}}-1>0 \\
& x\ne 0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \left[ \begin{aligned}
& x>1 \\
& x<-1 \\
\end{aligned} \right. \\
& x\ne 0 \\
\end{aligned} \right.$.
Tập xác định. $D=\left( -\infty ;-1 \right)\cup \left( 1;+\infty \right)$.
Ta có.
${f}'\left( x \right)=\dfrac{{{\left( 1-\dfrac{1}{{{x}^{2}}} \right)}^{\prime }}}{1-\dfrac{1}{{{x}^{2}}}}$ $=\dfrac{\dfrac{2}{{{x}^{3}}}}{1-\dfrac{1}{{{x}^{2}}}}=\dfrac{2}{\left( x-1 \right).x.\left( x+1 \right)}$ $=\dfrac{2}{x}.\dfrac{1}{2}.\left( \dfrac{1}{x-1}-\dfrac{1}{x+1} \right)$
$=\dfrac{1}{x}.\dfrac{1}{x-1}-\dfrac{1}{x}.\dfrac{1}{x+1}$ $=\dfrac{1}{x-1}-\dfrac{1}{x}-\dfrac{1}{x}+\dfrac{1}{x+1}$.
Do đó.
${f}'\left( 2 \right)=1-\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{3}$.
${f}'\left( 3 \right)=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{4}$.
${f}'\left( 4 \right)=\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{4}+\dfrac{1}{5}$.
…………………………………
${f}'\left( 2018 \right)=\dfrac{1}{2017}-\dfrac{1}{2018}-\dfrac{1}{2018}+\dfrac{1}{2019}$
${f}'\left( 2019 \right)=\dfrac{1}{2018}-\dfrac{1}{2019}-\dfrac{1}{2019}+\dfrac{1}{2020}$.
${f}'\left( 2020 \right)=\dfrac{1}{2019}-\dfrac{1}{2020}-\dfrac{1}{2020}+\dfrac{1}{2021}$.
$\Rightarrow {f}'\left( 2 \right)+{f}'\left( 3 \right)+...+{f}'\left( 2019 \right)+{f}'\left( 2020 \right)=1-\dfrac{1}{2}-\dfrac{1}{2020}+\dfrac{1}{2021}$ $=\dfrac{1010.2021-1}{2020.2021}$.
Suy ra $m=1010.2021-1$ ; $n=2020.2021$.
Vậy $S=2m-n=-2$.
& {{x}^{2}}-1>0 \\
& x\ne 0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \left[ \begin{aligned}
& x>1 \\
& x<-1 \\
\end{aligned} \right. \\
& x\ne 0 \\
\end{aligned} \right.$.
Tập xác định. $D=\left( -\infty ;-1 \right)\cup \left( 1;+\infty \right)$.
Ta có.
${f}'\left( x \right)=\dfrac{{{\left( 1-\dfrac{1}{{{x}^{2}}} \right)}^{\prime }}}{1-\dfrac{1}{{{x}^{2}}}}$ $=\dfrac{\dfrac{2}{{{x}^{3}}}}{1-\dfrac{1}{{{x}^{2}}}}=\dfrac{2}{\left( x-1 \right).x.\left( x+1 \right)}$ $=\dfrac{2}{x}.\dfrac{1}{2}.\left( \dfrac{1}{x-1}-\dfrac{1}{x+1} \right)$
$=\dfrac{1}{x}.\dfrac{1}{x-1}-\dfrac{1}{x}.\dfrac{1}{x+1}$ $=\dfrac{1}{x-1}-\dfrac{1}{x}-\dfrac{1}{x}+\dfrac{1}{x+1}$.
Do đó.
${f}'\left( 2 \right)=1-\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{3}$.
${f}'\left( 3 \right)=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{4}$.
${f}'\left( 4 \right)=\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{4}+\dfrac{1}{5}$.
…………………………………
${f}'\left( 2018 \right)=\dfrac{1}{2017}-\dfrac{1}{2018}-\dfrac{1}{2018}+\dfrac{1}{2019}$
${f}'\left( 2019 \right)=\dfrac{1}{2018}-\dfrac{1}{2019}-\dfrac{1}{2019}+\dfrac{1}{2020}$.
${f}'\left( 2020 \right)=\dfrac{1}{2019}-\dfrac{1}{2020}-\dfrac{1}{2020}+\dfrac{1}{2021}$.
$\Rightarrow {f}'\left( 2 \right)+{f}'\left( 3 \right)+...+{f}'\left( 2019 \right)+{f}'\left( 2020 \right)=1-\dfrac{1}{2}-\dfrac{1}{2020}+\dfrac{1}{2021}$ $=\dfrac{1010.2021-1}{2020.2021}$.
Suy ra $m=1010.2021-1$ ; $n=2020.2021$.
Vậy $S=2m-n=-2$.
Đáp án C.