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Câu hỏi: Cho hàm số $f\left( x \right)=\dfrac{x}{\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)...\left( x-2020 \right)}$. Giá trị của ${f}'\left( 0 \right)$ là
A. $-\dfrac{1}{2019!}$
B. $2020!$
C. $\dfrac{1}{2019!}$
D. $\dfrac{1}{2020!}$
A. $-\dfrac{1}{2019!}$
B. $2020!$
C. $\dfrac{1}{2019!}$
D. $\dfrac{1}{2020!}$
Đặt $g\left( x \right)=\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)...\left( x-2020 \right)$
Ta có $f\left( x \right)=\dfrac{x}{g\left( x \right)}$
${f}'\left( x \right)=\dfrac{g\left( x \right)-x{g}'\left( x \right)}{{{\left[ g\left( x \right) \right]}^{2}}}=\dfrac{1}{g\left( x \right)}-x.\dfrac{{g}'\left( x \right)}{{{\left[ g\left( x \right) \right]}^{2}}}\Rightarrow {f}'\left( 0 \right)=\dfrac{1}{g\left( 0 \right)}-0.\dfrac{{g}'\left( 0 \right)}{{{\left[ g\left( 0 \right) \right]}^{2}}}=\dfrac{1}{g\left( 0 \right)}$
$\Rightarrow {f}'\left( 0 \right)=\dfrac{1}{\left( -1 \right)\left( -2 \right)\left( -3 \right)...\left( -2020 \right)}=\dfrac{1}{2020!}$
Ta có $f\left( x \right)=\dfrac{x}{g\left( x \right)}$
${f}'\left( x \right)=\dfrac{g\left( x \right)-x{g}'\left( x \right)}{{{\left[ g\left( x \right) \right]}^{2}}}=\dfrac{1}{g\left( x \right)}-x.\dfrac{{g}'\left( x \right)}{{{\left[ g\left( x \right) \right]}^{2}}}\Rightarrow {f}'\left( 0 \right)=\dfrac{1}{g\left( 0 \right)}-0.\dfrac{{g}'\left( 0 \right)}{{{\left[ g\left( 0 \right) \right]}^{2}}}=\dfrac{1}{g\left( 0 \right)}$
$\Rightarrow {f}'\left( 0 \right)=\dfrac{1}{\left( -1 \right)\left( -2 \right)\left( -3 \right)...\left( -2020 \right)}=\dfrac{1}{2020!}$
Đáp án D.