a²＋2a=2 b²+2b=2 则b分之a+a分之b=_ （x+y）（x+y+2）-8=0 求x+ 10
a²＋2a=2b²+2b=2则b分之a+a分之b=_（x+y）（x+y+2）-8=0求x+y求学霸...
a²＋2a=2
b²+2b=2
则b分之a+a分之b=_
（x+y）（x+y+2）-8=0
求x+y
求学霸 展开
b²+2b=2
则b分之a+a分之b=_
（x+y）（x+y+2）-8=0
求x+y
求学霸 展开
1个回答
展开全部
consider
x^2+2x-2=0
sum of roots =a+b = -2
products of roots =ab =-2
a/b + b/a
= (a^2+b^2) / (ab)
=[(a+b)^2 -2ab]/(ab)
=(4+4)/(-2)
=-4
(x+y)(x+y+2) - 8 =0
(x+y)^2+ 2(x+y) - 8 =0
(x+y+4)(x+y-2)=0
x+y = -4 or 2
x^2+2x-2=0
sum of roots =a+b = -2
products of roots =ab =-2
a/b + b/a
= (a^2+b^2) / (ab)
=[(a+b)^2 -2ab]/(ab)
=(4+4)/(-2)
=-4
(x+y)(x+y+2) - 8 =0
(x+y)^2+ 2(x+y) - 8 =0
(x+y+4)(x+y-2)=0
x+y = -4 or 2
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
