Question

E(x)=(x-2008)$^{2009}$ + (2008-x)$^{2009}$

Answer 1

Notam : t = x - 2008
-t = -x + 2008 => -t = 2008-x
E(x) devine:
$E(x)=t^{2009}+(-t)^{2009}=t^{2009}-t^{2009}=\ \textgreater \ E(x)=0$