SOS 谁能帮我解数学题，谢谢了!!!

(x-3)^2+|3y+1|=0
(x-3)^2=0, x=3
|3y+1|=0. y=-1/3

3x^2y-[2xy^2-2(xy-3/2)+xy]+3xy^2
=3x^2y-[2xy^2-2xy+3/2+xy]+3xy^2
=3x^2y-[2xy^2-xy+3/2]+3xy^2
=3x^2y-2xy^2+xy-3/2+3xy^2
=x^2y+xy+3xy^2-3/2
=xy(x+1+3y)-3/2
=-3*1/3(3+1-3*1/3)-3/2
=-3-3/2
=-9/2

先解出X、 Y 因为（X-3)的方大于等于0 3Y+1的绝对值大于等于0 ，而和为0 ，
则只能同时为0 X-3=0 3Y+1=0联立即可 ，再求后面的

二分之九噻！具体请见首席运营官的慷慨陈词了啦！抱歉，小ZX!

用：x-3)^2+|3y+1|=0
(x-3)^2=0, x=3
|3y+1|=0. y=-1/3

3x^2y-[2xy^2-2(xy-3/2)+xy]+3xy^2
=3x^2y-[2xy^2-2xy+3/2+xy]+3xy^2
=3x^2y-[2xy^2-xy+3/2]+3xy^2
=3x^2y-2xy^2+xy-3/2+3xy^2
=x^2y+xy+3xy^2-3/2
=xy(x+1+3y)-3/2
=-3*1/3(3+1-3*1/3)-3/2
=-3-3/2
=-9/2
简简单单便做出来了啊！是：9/2=4.5