已知a-2的绝对值+(b+1)^2=0求3(2a^2b-ab^2)-(5a^2b-4ab^2)的值

|a-2|+(b+1)^2=0
两项均为非负值,和为0,则它们都是0
∴a-2=0,b+1=0
∴a=2,b=-1
∴3(2a^2b-ab^2)-(5a^2b-4ab^2)
=6a²-3ab²-5a²b+4ab²
=a²b+ab²
=ab(a+b)
=2×(-1)(2-1)
=-2