考點(diǎn):命題的真假判斷與應(yīng)用
專題:函數(shù)的性質(zhì)及應(yīng)用,不等式的解法及應(yīng)用
分析:A,利用不等式的性質(zhì),a>b>0⇒,a
2>ab;又a>c⇒ab>bc,于是可得a
2>bc,可判斷A;
B,舉例說明,-1>-2>-3,則
=
<
=
,可判斷B;
C,舉例說明,-2>-3,(-2)
2<(-3)
2,可判斷C;
D,利用對(duì)數(shù)函數(shù)的性質(zhì),可判斷D.
解答:
解:對(duì)于A,若a>b>0,則a
2>ab;又a>c,則ab>bc,故a
2>bc,A正確;
對(duì)于B,若a>b>c則
>,錯(cuò)誤.如-1>-2>-3,則
=
<
=
,故B錯(cuò)誤;
對(duì)于C,若a>b,n∈N
*則a
n>b
n,錯(cuò)誤,如-2>-3,(-2)
2<(-3)
2,故C錯(cuò)誤;
對(duì)于D,若a>b>0,則lna>lnb,故D錯(cuò)誤;
綜上所述,A、B、C、D四個(gè)選項(xiàng)中,只有A正確.
故選:A.
點(diǎn)評(píng):本題考查命題的真假判斷與應(yīng)用,著重考查不等式的性質(zhì)及應(yīng)用,屬于中檔題.