[tex]f(1-x)=\frac{9^{1-x}-2}{9^{1-x}+3}=\frac{9-2.9^x}{9+3.9^x}=\frac{9-2.9^x}{3(9^x+3)}[/tex]
[tex]\Rightarrow f(x)+f(1-x)=\frac{3(9^x-2)+9-2.9^x}{3(9^x+3)}=\frac{9^x+3}{3(9^x+3)}=\frac{1}{3}[/tex]
[tex]\Rightarrow P=f\left ( \frac{1}{2017} \right )+f\left ( \frac{2016}{2017} \right )+...+f\left ( \frac{1008}{2017} \right )+f\left ( \frac{1009}{2017} \right )+f(1)[/tex]
[tex]=\frac{1}{3}+\frac{1}{3}+...+\frac{1}{3}+\frac{7}{12}=\frac{4039}{12}[/tex]