Question

若(a+b+c)^2=3ab+3bc+3ac.求证a=b=c.

Answer 1

(a + b + c)² = 3ab + 3bc + 3ac
a² + b² + c² + 2ab + 2bc + 2ac = 3ab + 3bc + 3ac
a² + b² + c² - ab - bc - ac = 0
2a² + 2b² + 2c² - 2ab - 2bc - 2ac = 0
a² - 2ab + b² + b² - 2bc + c² + a² - 2ac + c² = 0
(a - b)² + (b - c)² + (a - c)² = 0
因为一个数的平方大于等于0
所以只有当 a - b = 0 且 b - c = 0 且 a - c = 0 时等号成立
所以 a = b = c

Answer 2

(a+b+c)^2=3ab+3bc+3ac
a2+b2+c2+2ab+2ac+2bc=3ab+3bc+3ac
a2+b2+c2-ab-ac-bc=0
方程两边同时乘以2
2（a2+b2+c2-ab-ac-bc）=0
（a-b）2+（b-c）2+（a-c）2=0
三个平方（大于等于0的数）相加，三个数各自为0，所以a-b=0，b-c=0，a-c=0，所以a=b=c

Answer 3

化简得:a^2+b^2+c^2=ab+bc+ac，两边同乘以2移项得:(a^2+b^2-2ab)+(a^2+c^2-2ac)+(b^2+c^2-2bc)=0→(a-b)^2+(a-c)^2+(b-c)^2=0→a=b=c

Answer 4

(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc
∵(a+b+c)^2=3ab+3bc+3ac
∴a^2+b^2+c^2=ab+ac+bc
∴a^2+b^2+c^2-ab-ac-bc=0
∴2(a^2+b^2+c^2-ab-ac-bc)=0
∴(a-b)^2+(a-c)^2-(b-c)^2=0
∴a=b=c