1.解:(1)1/2ab²•(2a²b-3ab²)=1/2ab²•(2a²b)+1/2ab²•(-3ab²)=a³b³-3/2a²
(2)x(2x-5)+3x(x+2)-5x(x-1)=2x²-5x+3x²+6x-5x²+5x=6x.
(3)a(a²+ab+b²)-b(a²+ab+b²)=a³+a²b+ab²-a²b-ab²-b³=a³-b³.
(4)原式=a³-3a+a³+3a²-3a³+3a=-a³+6a².
2.解:因为A=-2ab,B=4ab(a-b)=4a²b-4ab²,所以A•B=-2ab(4a²b-4ab²)=-8a³b²+8a²b³.
3.(1)-6ab (2)2a (3)2ab 5b² (4)1 4ab 8a²b²
4.解:(a+b)(c+d)=(a+b)•c+(a+b)•d=ac+bc+ad+bd.