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題目有誤:當k=__時,函數y=2x^(k^2-3k-3)為正比例函數. 因為是正比例函數, 所以k^2-3k-3=1, k^2-3k-4=0, (k-4)(k+1)=0 所以k1=4,k2=-1 所以當k=4或-1時,函數y=2x^(k^2-3k-3)為正比例函數
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