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