(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)=03(a-c)^2=0a-c=0why得不出来

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(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)=03(a-c)^2=0a-c=0why得不出来

(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)=03(a-c)^2=0a-c=0why得不出来
(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

(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)=03(a-c)^2=0a-c=0why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

要求证的等式没写清楚
(b-c)^2=(c-a)^2=(a-b)^2 则 a-b、a-c、b-c三个的绝对值相等
a-b绝对值=a-c的绝对值 则a-b=a-c或a-b=c-a
前者a=b=c,
后者2a=b+c再代入原绝对值等式中得1/2(b-c)绝对值=b-c的绝对值,所以b-c=0,所以a=b=c
综上得a=b=c

(b-c)^2=(c-a)^2=(a-b)^2
(b-c)^2=(c-a)^2
(b-c)^2-(c-a)^2=0
(b-c+c-a)(b-c-c+a)=0
(b-a)(b+a)=0
b^2-a^2=0
b^2=a^2
同理
(b-c)^2=(c-a)^2=(a-b)^2
(b-c)^2=(a-b)^2
(b-c)...

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(b-c)^2=(c-a)^2=(a-b)^2
(b-c)^2=(c-a)^2
(b-c)^2-(c-a)^2=0
(b-c+c-a)(b-c-c+a)=0
(b-a)(b+a)=0
b^2-a^2=0
b^2=a^2
同理
(b-c)^2=(c-a)^2=(a-b)^2
(b-c)^2=(a-b)^2
(b-c)^2-(a-b)^2=0
(b-c+a-b)(b-c-a+b)=0
(a-c)(-a-c)=0
-(a-c)(a+c)=0
a^2-c^2=0
a^2=c^2
b^2=a^2=c^2
则b=a=c
即a=b=c

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