解1题
原式=(1/2)a-2[a-(1/3)b²]+[(-3/2)ab+(1/3)b²]
=(1/2)a-2a+(2/3)b²-(3/2)ab+(1/3)b²
=[(1/2)a-2a]+[(2/3)b²+(1/3)b²]-(3/2)ab
=(-3/2)a+b²-(3/2)ab
解2题
原式=3x-y²+(4y²-xy)-2(x-xy)
=3x-y²+4y²-xy-2x+2xy
=(3x-2x)+(-y²+4y²)+(-xy+2xy)
=x+3y²+xy
当x=0.5,y=3时
原式=0.5+3×3²+0.5×3
=0.5+27+1.5
=29