题目应是：先化简,再求值：
若（x-3）^2+|y+2|=0,求代数式3x^2y-[xy^2-2（2xy^2-3x^2y）+x^2y]+4xy^2的值．
若（x-3）平方+|y+2|=0
x-3=0 y+2=0
x=3 y=-2
代数式3x^2y-[xy^2-2（2xy^2-3x^2y）+x^2y]+4xy^2的值
=3x^2y-[xy^2-4xy^2+6x^2y+x^2y]+4xy^2
=3x^2y-[-3xy^2+7x^2y]+4xy^2
=3x^2y+3xy^2-7x^2y+4xy^2
=-4x^2y+7xy^2
=-4*3^2*(-2)+7*3*(-2)^2
=72+84
=156