f(x+1)=x²-4
设x+1=t
x=t-1
f(t)=(t-1)²-4
=t²-2t-3
∴f(x)=x²-2x-3
f(x)≤0
∴x²-2x-3≤0
(x-3)(x+1)≤0
-1≤x≤3
解集是{x|-1≤x≤3}
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