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f(x+1)=loga(x-1)/(3+x)(a>0且a≠1) 令x+1=t 則x=t-1 ∴f(t)=loga [((t-2)/(t+2)] ∴f(x)=loga(x-2)/(x+2) 定義域是(x-2)/(x+2)>0 即x>2或x2或x
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