If f (x) = ax ^ 2008 + x ^ 2009 + BX ^ 2010-8 and f (1) = 10, then f (- 1)=

If f (x) = ax ^ 2008 + x ^ 2009 + BX ^ 2010-8 and f (1) = 10, then f (- 1)=

Because f (1) = 10, substitute x = 1, f (x) = 10 to get:
a+1+b-8=10
a+b=17
f(-1)=a-1+b-8=17-1-8=8

Equation (x2006 + 1) (1 + x2 + X4 +...) +The number of real number solutions of (x2004) = 2006x2005 is______ ．

（x2006+1）（1+x2+x4+… +X2004) = 2006x2005, equivalent to (x + 1x2005) (1 + x2 + X4 + +X2004) = 2006 is equivalent to x + X3 + X5 + +x2005+1x2005+1x2003+1x2001+… +So 2006 = x + 1x + X3 + 1x3 + +X2005 + 1x2005 ≥ 2 × 1003 = 2006. The equal sign holds if and only if x = 1. So x = 1 is all the solutions of the original equation. Therefore, the number of real number solutions of the original equation is 1, so the answer is 1

If 1 + X + x2 + X3 = 0, then x + x2 + X3 + +The value of x2004 is______ ．

∵1+x+x2+x3=0，∴x+x2+x3+… +x2004，=x（1+x+x2+x3）+x5（1+x+x2+x3）+x9（1+x+x2+x3）+… +x1997（1+x+x2+x3）+x2001（1+x+x2+x3），=（1+x+x2+x3）（x+x5+x9+x12+… +So the answer is 0