已知(a+3)²+|a+b+5|=0,求3a²b-[2a²b-（2ab-a²b）-4b²]-ab的值

解
(a+3)²+/a+b+5/=0
∴a+3=0,a+b+5=0
∴a=-3.b=-2
∴原式
=3a²b-(2a²b-2ab+a²b-4b²)-ab
=3a²b-(3a²b-2ab-4b²)-ab
=2ab+4b²-ab
=ab+4b²
=-3×(-2)+4(-2)²
=6+16
=22