先化简再求值:x2-2xy+y2分之x2+xy-2y2-(1-xy2)(1+xy2)-x2÷y的四次方分之一,其中x=根3-根2 y=根3-根2
1个回答
展开全部
解
(x²+xy-2y²)/(x²-2xy+y²)-(1-xy²)(1+xy)²-x²÷(1/y^4)——这个吗
=(x+2y)(x-y)/(x-y)²-(1-x²y的4次方)-x²y的4次方
=(x+2y)/(x-y)-1+x²y的4次方-x²y的4次方
=(x+2y)/(x-y)-1
=(x+2y)/(x-y)-(x-y)/(x-y)
=(x+2y-x+y)/(x-y)
=(3y)/(x-y)——x≠y不对啊
(x²+xy-2y²)/(x²-2xy+y²)-(1-xy²)(1+xy)²-x²÷(1/y^4)——这个吗
=(x+2y)(x-y)/(x-y)²-(1-x²y的4次方)-x²y的4次方
=(x+2y)/(x-y)-1+x²y的4次方-x²y的4次方
=(x+2y)/(x-y)-1
=(x+2y)/(x-y)-(x-y)/(x-y)
=(x+2y-x+y)/(x-y)
=(3y)/(x-y)——x≠y不对啊
本回答由提问者推荐
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
