Question

已知a－b=3,b－c=﹣2,求a²＋b²＋c²－ab－bc－ca的值

Answer 1

a-b=3
b-c=-2
相加得
a-c=1

a²＋b²＋c²－ab－bc－ca
=[(a-b)²+(a-c)²+(b-c)²]/2
=[3²+(-2)²+1]/2
=(9+4+1)/2
=7

Answer 2

a－b=3,
b－c=﹣2,
两式相加，得
a-c=1
a²＋b²＋c²－ab－bc－ca
=（2a²＋2b²＋2c²－2ab－2bc－2ca）/2
=[（a²-2ab+b²）+（a²-2ac+c²）+（b²-2bc+c²）]/2
=[（a-b）²+（a-c）²+（b-c）²]/2
=[3²+1²+（-2）²]/2
=14/2
=7