a²+b²+c²-ab-bc-ca
=(1/2)(2a²+2b²+2c²-2ab-2bc-2ca)
=(1/2)[(a-b)^2+(b-c)^2+(a-c)^2]
=(1/2)[3^2+1^2+4^2]
=13
根据上面那个关系
a²+b²+c²-ab-bc-ca
=(1/2)[(a-b)^2+(b-c)^2+(a-c)^2]
=(1/2)[2^2+2^2+4^2]
=12
所以a²+b²+c²=17
a²+b²+c²-ab-bc-ca
=(1/2)(2a²+2b²+2c²-2ab-2bc-2ca)
=(1/2)[(a-b)^2+(b-c)^2+(a-c)^2]
=(1/2)[3^2+1^2+4^2]
=13
根据上面那个关系
a²+b²+c²-ab-bc-ca
=(1/2)[(a-b)^2+(b-c)^2+(a-c)^2]
=(1/2)[2^2+2^2+4^2]
=12
所以a²+b²+c²=17