已知（a+2）的2014次方+（a+b+5）的2012次方=0,求3a²b-[2a²b-(2ab-a

a+2=0
a+b+5=0
a=-2
b=-3
3a²b-[2a²b-(2ab-a²b)-4a²]-ab
=3a²b-2a²b+2ab+a²b+4a²-ab
=4a²+2a²b+ab
=4*4+2*4*(-3)+2*3
=16-24+6
=-2