Cho \(f(x) = \dfrac{{{{2018}^x}}}{{{{2018}^x} + \s...

A

- Hướng dẫn giải

Phương pháp giải:

Tính f(x) và f(1-x) và tìm mối liên hệ giữa f(x) và f(1-x)

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.