这个式子是怎么化简出来的
x/(x^2+1)-y/(y^2+1)=[(x-y)(1-xy)]/[(x^2+1)(y^2+1)]我算了很多次我只能化成[(x-y)(x^2+y^2+1)]/[(x^2...
x/(x^2+1)-y/(y^2+1)=[(x-y)(1-xy)]/[(x^2+1)(y^2+1)] 我算了很多次我只能化成[(x-y)(x^2+y^2+1)]/[(x^2+1)(y^2+1)] 而已
展开
1个回答
展开全部
=[y^2*x+x-y-x^2*y]/[(x^2+1)(y^2+1)]
=[(x-y)+xy(y-x)}/[(x^2+1)(y^2+1)]
=[(x-y)-xy(x-y)]/[(x^2+1)(y^2+1)]
=[(x-y)(1-xy)]/[(x^2+1)(y^2+1)]
=[(x-y)+xy(y-x)}/[(x^2+1)(y^2+1)]
=[(x-y)-xy(x-y)]/[(x^2+1)(y^2+1)]
=[(x-y)(1-xy)]/[(x^2+1)(y^2+1)]
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
