Đáp án đúng: D
Giải chi tiết:Ta có: \(f\left( x \right) = \dfrac{x}{{{{2020}^x} + 1}}\)
\(\begin{array}{l} \Rightarrow f\left( { - x} \right) = \dfrac{{ - x}}{{{{2020}^{ - x}} + 1}} = \dfrac{{ - x}}{{\dfrac{1}{{{{2020}^x}}} + 1}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{ - x}}{{\dfrac{{{{2020}^x} + 1}}{{{{2020}^x}}}}} = \dfrac{{ - x{{.2020}^x}}}{{{{2020}^x} + 1}}\end{array}\)
\(\begin{array}{l} \Rightarrow f\left( x \right) - f\left( { - x} \right)\\ = \dfrac{x}{{{{2020}^x} + 1}} + \dfrac{{x{{.2020}^x}}}{{{{2020}^x} + 1}}\\ = \dfrac{{x + x{{.2020}^x}}}{{{{2020}^x} + 1}} = \dfrac{{x\left( {{{2020}^x} + 1} \right)}}{{{{2020}^x} + 1}} = x\end{array}\)
Khi đó ta có:
\(\begin{array}{l}{S_1} - {S_2} = \left[ {f\left( 1 \right) + f\left( 2 \right) + ... + f\left( {100} \right)} \right] - \left[ {f\left( { - 1} \right) + f\left( { - 2} \right) + ... + f\left( { - 100} \right)} \right]\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {f\left( 1 \right) - f\left( { - 1} \right)} \right] + \left[ {f\left( 2 \right) - f\left( { - 2} \right)} \right] + ... + \left[ {f\left( {100} \right) - f\left( { - 100} \right)} \right]\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 + 2 + ... + 100\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{100.101}}{2} = 5050\end{array}\)
Chọn D.
