Cho hàm số \(f\left( x \right) = \frac{{{4^x}}}{{{...

B

- Hướng dẫn giải

Nhận thấy: \(f\left( x \right) + f\left( {x - 1} \right) = \frac{{{4^x}}}{{{4^x} + 2}} + \frac{{{4^{1 - x}}}}{{{4^{1 - x}} + 2}} = \frac{{{4^x}}}{{{4^x} + 2}} + \frac{{\frac{4}{{{4^x}}}}}{{\frac{4}{{{4^x}}} + 2}} = \frac{{{4^x}}}{{{4^x} + 2}} + \frac{2}{{{4^x} + 2}} = 1\)

\( \Rightarrow S = \left[ {f\left( {\frac{1}{{2019}}} \right) + f\left( {\frac{{2018}}{{2019}}} \right)} \right] + \left[ {f\left( {\frac{2}{{2019}}} \right) + f\left( {\frac{{2017}}{{2019}}} \right)} \right] + ... + \left[ {f\left( {\frac{{1009}}{{2019}}} \right) + f\left( {\frac{{1010}}{{2019}}} \right)} \right] = 1009\)