Đáp án đúng: A
Giải chi tiết:\(\begin{array}{l}f(x) = \dfrac{{{{2018}^x}}}{{{{2018}^x} + \sqrt {2018} }}\\f(1 - x) = \dfrac{{{{2018}^{1 - x}}}}{{{{2018}^{1 - x}} + \sqrt {2018} }} = \dfrac{{\dfrac{{2018}}{{{{2018}^x}}}}}{{\dfrac{{2018}}{{{{2018}^x}}} + \sqrt {2018} }} = \dfrac{{2018}}{{2018 + \sqrt {2018} {{.2018}^x}}} = \dfrac{{\sqrt {2018} }}{{\sqrt {2018} + {{2018}^x}}}\\f(x) + f(1 - x) = \dfrac{{{{2018}^x}}}{{{{2018}^x} + \sqrt {2018} }} + \dfrac{{\sqrt {2018} }}{{\sqrt {2018} + {{2018}^x}}} = 1\end{array}\)
Ta có :
\(\begin{array}{l}S = f\left( {\dfrac{1}{{2017}}} \right) + f\left( {\dfrac{2}{{2017}}} \right) + ... + f\left( {\dfrac{{2016}}{{2017}}} \right)\\ = \left[ {f\left( {\dfrac{1}{{2017}}} \right) + f\left( {\dfrac{{2016}}{{2017}}} \right)} \right] + \left[ {f\left( {\dfrac{2}{{2017}}} \right) + f\left( {\dfrac{{2015}}{{2017}}} \right)} \right] + ... + \left[ {f\left( {\dfrac{{1008}}{{2017}}} \right) + f\left( {\dfrac{{1009}}{{2017}}} \right)} \right]\\ = \left[ {f\left( {\dfrac{1}{{2017}}} \right) + f\left( {1 - \dfrac{1}{{2017}}} \right)} \right] + \left[ {f\left( {\dfrac{2}{{2017}}} \right) + f\left( {1 - \dfrac{2}{{2017}}} \right)} \right] + ... + \left[ {f\left( {\dfrac{{1008}}{{2017}}} \right) + f\left( {1 - \dfrac{{1008}}{{2017}}} \right)} \right]\\ = \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ... + \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\end{array}\)
( có: 1008 số 1)
= 1.1008 = 1008.
Chọn: A.
