函数f(x)满足：①定义域为R；②对任意的x∈R,有f(x+2)=-2f(x)；③当x∈[-1,1]时f(x)=-｜x｜+1,

当x∈[-1,1]时f(x)=-｜x｜+1,所以f(-1)=f(1)=0,f(0)=1
又对任意的x∈R,有f(x+2)=-2f(x),
所以f(-9)=f(-7)=f(-5)=f(-3)=f(-1)=f(1)=t((3)=f(5)=f(7)=f(9)=0,
f(2)=-2f(0)=-2,f(4)=-2f(2)=4,f(6)=-8,f(8)=16,f(10)=-32,
f(-2)=-1/2,f(-4)=1/4,f(-6)=-1/8,f(-8)=1/16,f(-10)=-1/32,
显然当x=±1时,f(±1)=log2(|±1|)=0,x=±1是方程f(x)=log2(|x|)的解.
当x>0 (或x