计算:(x的-1次方-y的-1次方)÷(x的-2次方-y的-2次方)-xy(x+y)的-1次方
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计算:(x的-1次方-y的-1次方)÷(x的-2次方-y的-2次方)-xy(x+y)的-1次方
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)
再问: 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。
再答: 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
=(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)
再问: 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。
再答: 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
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