原等式可变为:
(x-y+y-z)^2-4(x-y)(y-z)=0 =>(x-y)^2+2(x-y)(y-z)-4(x-y)(y-z)+(y-z)^2=0
=>(x-y-y+z)^2=0 由实数平方大于等于0可知:x+z=2y
如果觉得满意我的回答,就采纳我一下吧~^-^
原等式可变为:
(x-y+y-z)^2-4(x-y)(y-z)=0 =>(x-y)^2+2(x-y)(y-z)-4(x-y)(y-z)+(y-z)^2=0
=>(x-y-y+z)^2=0 由实数平方大于等于0可知:x+z=2y
如果觉得满意我的回答,就采纳我一下吧~^-^