解:(1)原式=-
ab-
a
2+
a
2+
ab=
ab+
a
2;
(2)原式=3a
2-[8a-4a+7-2a
2]=3a
2-4a-7+2a
2=5a2-4a-7;
(3)2A-B=2(x
2+5x)-(3x
2+2x-6)=2x
2+10x-3x
2-2x+6=-x2+8x+6;
當(dāng)x=-3時(shí),原式=-(-3)
2+8×(-3)+6=-9-24+6=-27;
(4)如右圖可知,
a>b>0,c<0,
∴|a-b|-|-b+c|+|c-a|=a-b-(b-c)+(-c+a)=a-b-b+c-c+a=2a-2b.
分析:(1)先去括號(hào),再合并即可;
(2)去括號(hào)后再進(jìn)行合并即可;
(3)先把A、B的值整體代入,再去括號(hào),合并同類項(xiàng),最后把x的值代入計(jì)算即可;
(4)先根據(jù)數(shù)軸得出a、b、c的取值范圍,再根據(jù)絕對(duì)值的定義進(jìn)行化簡即可.
點(diǎn)評(píng):本題考查了整式的化簡以及求值、絕對(duì)值的概念,解題的關(guān)鍵是去括號(hào)法則和合并同類項(xiàng),去掉絕對(duì)值后要保證結(jié)果是非負(fù)數(shù).