分式性质的数学题？~~~不会做

xy/x+y=1/2可以得到x+y=2xy
代入(3x-xy+3y)/(x-xy+y)=[3(x+y)-xy]/[(x+y)-xy]=(6xy-xy)/(2xy-xy)=5xy/xy=5

xy/x+y=1/2
xy=1/2(x+y)
2xy=x+y
(3x-xy+3y)/(x-xy+y)=(3x+3y-xy)/(x+y-xy)=[3(x+y)-xy]/(x+y-xy)=(6xy-xy)/(2xy-xy)=5xy/xy
因为x不等于零,y不等于零,所以xy不等于零
所以5xy/xy=5

xy/x+y=1/2推出xy = (x+y)/2
3x-xy+3y/x-xy+y
= [3(x+y) - xy]/[(x+y)-xy]
= [3(x+y) - (x+y)/2]/[(x+y)-(x+y)/2]
= (5/2)/(1/2)
= 5

xy/x+y=1/2得到：x＋y＝2xy
将x＋y＝2xy代入3x-xy+3y/x-xy+y＝3（x＋y)-xy/x+y-xy
将x＋y＝2xy代入得:6xy-xy/2xy-xy=5

已知xy/x+y=1/2，求分式3x-xy+3y/x-xy+y
∵xy/x+y=1/2
xy=(x+y)/2
∴3x-xy+3y/x-xy+y
=[3(x+y)-xy]/[(x+y)-xy]
=[3(x+y)-(x+y)/2]/[(x+y)-(x+y)/2]
=[5(x+y)/2]/[(x+y)/2]
=5