已知a+10=b+7=c+6,(1)求（a-b）²+（b-c）²的值；（2）求代数式a²+b²+c²-ab-bc-a

Question

已知a+10=b+7=c+6,(1)求（a-b）²+（b-c）²的值；（2）求代数式a²+b²+c²-ab-bc-a

Answer 1

因为 a + 10 = b + 7 = c + 6
所以 a - b = -3 ,a - c = -4 ,b - c = -1
(1)
(a - b)² + (b - c)²
= (-3)² + (-1)²
= 9 + 1
= 10
(2)
a² + b² + c² - ab - bc - ac
= (1/2)×[(a² - 2ab + b²) + (a² - 2ac + c²) + (b² - 2bc + c²)]
= (1/2)×[(a - b)² + (a - c)² + (b - c)²]
= (1/2)×[(-3)² + (-4)² + (-1)²]
= (1/2)×(9 + 16 + 1)
= 13