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f'(x)=3mx²;-3(m+1)x+n =>3m-6m-6+n=0 =>n-3m=6 =>n=3(m+2) 已知x=1是f(x)=mx^3-3(m+1)x^2+nx+1的一個極值點,其中m,n屬於R,m
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