(b-c)^2=(c-a)^2=(a-b)^2 求证:a=b=c (b-c)^2-(c-a)^2=0 (b-c+c-a)(b-c-c+ why

问题描述:

(b-c)^2=(c-a)^2=(a-b)^2 求证:a=b=c (b-c)^2-(c-a)^2=0 (b-c+c-a)(b-c-c+ why
(a-c)(3c-3a)=0
3(a-c)^2=0
a-c=0
why得不出来

(a-c)[-3(a-c)]=0
所以-3(a-c)²=0
即3(a-c)²=0
a-c=0(c-a)^2-(2a-2c)^2=0(c-a+2a-2c)(c-a-2a+2c)=0why平方差得不出来那就算了(c-a+2a-2c)(c-a-2a+2c)=0为什么是× 呢我看还是你的好x²-y²=(x+y)(x-y)这都不是知道??行了,采纳吧