Đáp án:
$\dfrac{4}{5}+1008\\$
Giải thích các bước giải:
$P=f(x)+f(1-x)=\dfrac{16^x}{16^x+4}+\dfrac{16^{1-x}}{16^{1-x}+4}\\
\rightarrow P=\dfrac{4^{2x}}{4^{2x}+4}+\dfrac{4^{2-2x}}{4^{2-2x}+4}\\
\rightarrow P=\dfrac{4^{2x-1}}{4^{2x-1}+1}+\dfrac{4^{1-2x}}{4^{1-2x}+1}\\
\rightarrow P=\dfrac{4^{2x-1}}{4^{2x-1}+1}+\dfrac{1}{4^{2x-1}+1}\\
\rightarrow P=1\\
\Rightarrow S=f(\dfrac{1}{2017})+f(\dfrac{2}{2017})+....+f(\dfrac{2016}{2017})+f(\dfrac{2017}{2017})\\
\rightarrow S=f(\dfrac{2017}{2017})+f(\dfrac{2016}{2017})+....+f(\dfrac{1}{2017})\\
\rightarrow 2S=f(\dfrac{2017}{2017})+(f(\dfrac{1}{2017})+f(\dfrac{2016}{2017}))+(f(\dfrac{2}{2017})+(\dfrac{2015}{2017}))+...+(f(\dfrac{2016}{2017})+f(\dfrac{1}{2017}))+f(\dfrac{2017}{2017})\\
\rightarrow 2S=2f(1)+2016\\
\rightarrow 2S=\dfrac{8}{5}+2016\\
\rightarrow S=\dfrac{4}{5}+1008\\$
