(x-4xy/x+y+y)÷(x/x+y-y/y-x-2xy/x^2-y^2)
1个回答
展开全部
(x-4xy/x+y+y)÷(x/x+y-y/y-x-2xy/x^2-y^2)
=[(x(x+y)-4xy+y(x+y))/(x+y)]÷(x/x+y +y/x-y -2xy/x^2-y^2)
=[(x^2+xy-4xy+xy+y^2)/(x+y)]÷[(x(x-y)+y(x+y) -2xy)/(x^2-y^2)]
=[(x^2-2xy+y^2)/(x+y)]÷[(x^2-xy+xy+y^2 -2xy)/(x^2-y^2)]
=[(x-y)^2/(x+y)]÷[(x^2-2xy+y^2)/(x^2-y^2)]
=[(x-y)^2/(x+y)]÷[(x-y)^2/(x-y)(x+y)]
=[(x-y)^2/(x+y)]÷[(x-y)/(x+y)]
=[(x-y)^2/(x+y)] *[(x+y)/(x-y)]
=x-y
=[(x(x+y)-4xy+y(x+y))/(x+y)]÷(x/x+y +y/x-y -2xy/x^2-y^2)
=[(x^2+xy-4xy+xy+y^2)/(x+y)]÷[(x(x-y)+y(x+y) -2xy)/(x^2-y^2)]
=[(x^2-2xy+y^2)/(x+y)]÷[(x^2-xy+xy+y^2 -2xy)/(x^2-y^2)]
=[(x-y)^2/(x+y)]÷[(x^2-2xy+y^2)/(x^2-y^2)]
=[(x-y)^2/(x+y)]÷[(x-y)^2/(x-y)(x+y)]
=[(x-y)^2/(x+y)]÷[(x-y)/(x+y)]
=[(x-y)^2/(x+y)] *[(x+y)/(x-y)]
=x-y
本回答由网友推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
