解
-3x²+ax-bx²-x+3
=(-3-b)x²+(a-1)x+3
与x的取值无关
则-3-b=0,a-1=0
∴a=1,b=-3
∴3a²b-[2ab²-2(ab-3/2a²b)+ab]+3ab
=3a²b-(2ab²-2ab+3a²b+ab)+3ab
=3a²b-(2ab²-ab+3a²b)+3ab
=3a²b-3a²b-2ab²+ab+3ab
=4ab-2ab²
=4×1×(-3)-2×1×(-3)²
=-12-18
=-30