已知f(x)=﹛(x-1)²,(x>0) f(x+1)+1,(x≦0)则f(1)+f(-1)的值为已知f(x)=﹛(x-1)²,(x>0)f(x+1)+1,(x≦0)则f(1)+f(-1)的值为
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已知f(x)=﹛(x-1)²,(x>0) f(x+1)+1,(x≦0)则f(1)+f(-1)的值为已知f(x)=﹛(x-1)²,(x>0)f(x+1)+1,(x≦0)则f(1)+f(-1)的值为
已知f(x)=﹛(x-1)²,(x>0) f(x+1)+1,(x≦0)则f(1)+f(-1)的值为
已知f(x)=﹛(x-1)²,(x>0)
f(x+1)+1,(x≦0)则f(1)+f(-1)的值为
已知f(x)=﹛(x-1)²,(x>0) f(x+1)+1,(x≦0)则f(1)+f(-1)的值为已知f(x)=﹛(x-1)²,(x>0)f(x+1)+1,(x≦0)则f(1)+f(-1)的值为
1>0 f(1)=(1-1)²=0
-1
解:
f(1)+f(-1)
=0+f(0)+1
=f(1)+1+1
=0+1+1
=2
答:……