原式=(1/x-1/y)÷(1/x²-1/y²)-xy(x+y)
=(1/x-1/y)÷[(1/x-1/y)(1/x+1/y)]-xy(x+y)
=1/(1/x+1/y)-xy(x+y)
=1/[(x+y)/xy]-xy(x+y)
=xy/(x+y)-xy(x+y)
=xy*[1-(x+y)²]/(x+y)
=xy(1-x-y)(1+x+y)/(x+y)原式=(1/x-1/y)÷(1/x²-1/y²)-xy(x+y)最后题目-xy(x+y)的-1次方，-1次方漏了，答案为0，可是做了几遍没有做出来，请求帮助，谢了。原式=(1/x-1/y)÷(1/x²-1/y²)-xy/(x+y)=(1/x-1/y)÷[(1/x-1/y)(1/x+1/y)]-xy/(x+y)=1/(1/x+1/y)-xy/(x+y)=1/[(x+y)/xy]-xy/(x+y)=xy/(x+y)-xy/(x+y)=0不懂可追问,有帮助请采纳,祝你学习进步,谢谢