Question

若a,b是不等的正數(shù),則ab^k+a^kb-(a^k+1+b^k+1)(k∈N^*)的符號(hào)是(    )

A.恒正                                     B.與k的奇偶有關(guān)

C.恒負(fù)                                     D.與a、b的大小有關(guān)

Answer 1

解析：ab^k+a^kb-(a^k+1+b^k+1)=a^k(b-a)+b^k(a-b)=(b-a)(a^k-b^k),

當(dāng)a＞b時(shí),有b-a＜0,a^k-b^k＞0,

則(a-b)(a^k-b^k)＜0.

當(dāng)a＜b時(shí),b-a＞0,a^k-b^k＜0,則(a-b)(a^k-b^k)＜0.

∴ab^k+a^kb-(a^k+1+b^k+1)＜0.

答案：C