解:(1)a
2-3a+8-3a
2+4a-6=-2a
2+a+2;
(2)∵A=5x+3y-2,B=2x-2y+3,
∴A-B
=(5x+3y-2)-(2x-2y+3)
=5x+3y-2-2x+2y-3
=3x+5y-5;
(3)∵(a+2)
2+|b-

|=0,∴a=-2,b=

,
則5(3a
2b-ab
2)-4(-ab
2+3a
2b)
=15a
2b-5ab
2+4ab
2-12a
2b
=3a
2b-ab
2
=3×(-2)
2×

-(-2)×(

)
2
=6.5.
分析:(1)合并同類項(xiàng)即可得到結(jié)果;
(2)將A與B代入A-B中,去括號(hào)合并即可得到結(jié)果;
(3)由兩個(gè)非負(fù)數(shù)之和為0,得到兩非負(fù)數(shù)分別為0求出a與b的值,將所求式子去括號(hào)合并得到最簡(jiǎn)結(jié)果,把a(bǔ)與b的值代入計(jì)算,即可求出值.
點(diǎn)評(píng):此題考查了整式的加減-化簡(jiǎn)求值,以及非負(fù)數(shù)的性質(zhì):偶次冪與絕對(duì)值,涉及的知識(shí)有:去括號(hào)法則,以及合并同類項(xiàng)法則,熟練掌握法則是解本題的關(guān)鍵.