Đáp án đúng: D
Giải chi tiết:\(\begin{array}{l}f(1 - x) = \frac{1}{2}{\log ^2}\left( {\frac{{2(1 - x)}}{{1 - (1 - x)}}} \right) = \frac{1}{2}{\log ^2}\left( {\frac{{2 - 2x}}{x}} \right)\\f(x) + f(1 - x) = \frac{1}{2}\left( {{{\log }_2}\left( {\frac{{2x}}{{1 - x}}} \right) + {{\log }_2}\left( {\frac{{2 - 2x}}{x}} \right)} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}.{\log _2}\left( {\frac{{2x}}{{1 - x}}.\frac{{2 - 2x}}{x}} \right) = \frac{1}{2}{\log _2}4 = 1\end{array}\)
Vậy \(f\left( x \right) + f\left( {1 - x} \right) = 1\)
\(\begin{array}{l}f\left( {\frac{1}{{2017}}} \right) + f\left( {\frac{{2016}}{{2017}}} \right) = 1\\\,f\left( {\frac{2}{{2017}}} \right) + f\left( {\frac{{2015}}{{2017}}} \right) = 1\end{array}\)
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Ta có: \(S = f\left( {\frac{1}{{2017}}} \right) + f\left( {\frac{2}{{2017}}} \right) + f\left( {\frac{3}{{2017}}} \right) + ... + f\left( {\frac{{2015}}{{2017}}} \right) + f\left( {\frac{{2016}}{{2017}}} \right).\)
\(\underbrace { = \left[ {f\left( {\frac{1}{{2017}}} \right) + f\left( {\frac{{2016}}{{2017}}} \right)} \right] + \left[ {f\left( {\frac{2}{{2017}}} \right) + f\left( {\frac{{2015}}{{2017}}} \right)} \right] + .... + \left[ {f\left( {\frac{{1008}}{{2017}}} \right) + f\left( {\frac{{1009}}{{2017}}} \right)} \right]}_{}\)
Tổng cộng có 1008 cặp, mà mỗi cặp lại có tổng bằng 1
\( \Rightarrow S = 1008.1 = 1008.\)
Chọn D.