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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. $-4$.
Tập xác định: $D=\left( -\infty ;-1 \right)\cup \left( 1;\infty \right)$
Ta có: $f\left( x \right)=\ln \left( 1-\dfrac{1}{{{x}^{2}}} \right)\Rightarrow {f}'\left( x \right)=\dfrac{2}{x\left( {{x}^{2}}-1 \right)}=\dfrac{2}{x\left( x-1 \right)\left( x+1 \right)}=\dfrac{1}{x\left( x-1 \right)}-\dfrac{1}{x\left( x+1 \right)}$
Khi đó: ${f}'\left( 2 \right)+{f}'\left( 3 \right)+...+{f}'\left( 2019 \right)+{f}'\left( 2020 \right)$
$=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{2019.2020}-\dfrac{1}{2020.2021}$ $=\dfrac{1}{1.2}-\dfrac{1}{2020.2021}$
$=\dfrac{1010.2021-1}{2020.2021}$.
Vậy $m=1010.2021-1; n=2020.2021$, suy ra $2m-n=-2$.
Ta có: $f\left( x \right)=\ln \left( 1-\dfrac{1}{{{x}^{2}}} \right)\Rightarrow {f}'\left( x \right)=\dfrac{2}{x\left( {{x}^{2}}-1 \right)}=\dfrac{2}{x\left( x-1 \right)\left( x+1 \right)}=\dfrac{1}{x\left( x-1 \right)}-\dfrac{1}{x\left( x+1 \right)}$
Khi đó: ${f}'\left( 2 \right)+{f}'\left( 3 \right)+...+{f}'\left( 2019 \right)+{f}'\left( 2020 \right)$
$=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{2019.2020}-\dfrac{1}{2020.2021}$ $=\dfrac{1}{1.2}-\dfrac{1}{2020.2021}$
$=\dfrac{1010.2021-1}{2020.2021}$.
Vậy $m=1010.2021-1; n=2020.2021$, suy ra $2m-n=-2$.
Đáp án C.