分析:對(duì)于A,當(dāng)0<x<1時(shí),lgx<0,
<0;對(duì)于B,
x-在(0,2]上單調(diào)增,所以x=2時(shí),
x-取得最大值;對(duì)于C,
x+在[2,+∞)上單調(diào)增,所以x=2時(shí),
x+的最小值為
2;對(duì)于D,當(dāng)x>0時(shí),
+≥2,當(dāng)且僅當(dāng)x=1時(shí),等號(hào)成立,故可判斷.
解答:解:對(duì)于A,當(dāng)0<x<1時(shí),lgx<0,
<0,結(jié)論不成立;
對(duì)于B,
x-在(0,2]上單調(diào)增,所以x=2時(shí),
x-取得最大值,故B不成立;
對(duì)于C,
x+在[2,+∞)上單調(diào)增,所以x=2時(shí),
x+的最小值為
2,故C錯(cuò)誤;
對(duì)于D,當(dāng)x>0時(shí),
+≥2,當(dāng)且僅當(dāng)x=1時(shí),等號(hào)成立,故D成立
故選D.
點(diǎn)評(píng):本題考查的重點(diǎn)是基本不等式的運(yùn)用,解題的關(guān)鍵是明確基本不等式的使用條件,屬于基礎(chǔ)題.