一道初二数学题!!!

2x-3xy+2y/x+2xy+y
=[(2x-3xy+2y)/xy]/[(x+2xy+y)/xy]
(利用分式的基本性质，分子、分母同时除以xy).
＝〔(2/y)-3+(2/x)]/[(1/y)+2+(1/x)]
=(10-3)/(5+2)
=1

2x-3xy+2y/x+2xy+y
=[(2x-3xy+2y)/xy]/[(x+2xy+y)/xy]
(利用分式的基本性质，分子、分母同时除以xy).
＝〔(2/y)-3+(2/x)]/[(1/y)+2+(1/x)]
=(10-3)/(5+2)
=1

∵1/x+1/y=5
∴xy≠0
利用分式的基本性质，分子、分母同时除以xy≠0
2x-3xy+2y/x+2xy+y
=[(2x-3xy+2y)/xy]/[(x+2xy+y)/xy]
＝〔(2/y)-3+(2/x)]/[(1/y)+2+(1/x)]
=(10-3)/(5+2)
=1

原式=2-7xy/(x+2xy+y)
7xy/(x+2xy+y)的倒数化简为 1/7y+2/7+1/7x
1/7y+1/7x=5/7，所以7xy/(x+2xy+y)=1，原式=1
：）

随便取几个值试试;一般这种题答案都是1