Question

已知a-b=2，b-c=3，求a^2+b^2+c^2-ab-bc-ca的值

Answer 1

a-b=2
b-c=3
相加
a-c=5

所以原式=(2a²+2b²+2c²-2ab-2bc-2ac)/2
=[(a²-2ab+b²)+(b²-2bc+c²)+(c²-2ac+a²)]/2
=[(a-b)²+(b-c)²+(a-c)²]/2
=(4+9+25)
=19

Answer 2

a²+b²+c²-ab-bc-ca
=(1/2)(2a²+2b²+2c²-2ab-2bc-2ca)
=(1/2)[a²-2ab+b²+b²-2bc+c²+c²-2ca+a²]
=(1/2)[(a-b)²+(b-c)²+(c-a)²]
=(1/2)[2²+3²+(c-a)²]
已知条件应该还有个c-a的值吧。