Question

已知a-b=10;b-c=5,求a²+b²+c²-ab-bc-ca的值

求做题过程

Answer 1

a-b=10...............1
b-c=5.......................2
1式+2式得
a-c=15

a²+b²+c²-ab-bc-ca
=1/2*(2a²+2b²+2c²-2ab-2bc-2ca)
=1/2*(a²-2ab+b²+b²-2bc+c²+a²-2ca+c²)
=1/2*[(a-b)²+(b-c)²+(a-c)²]
=1/2*[10²+5²+15²]
=1/2*[100+25+225]
=1/2*350
=175

Answer 2

由a-b=10
b-c=5
可得a-c=15
a²+b²+c²-ab-bc-ca=（2a²+2b²+2c²-2ab-2bc-2ca）/2
=【（a-b）²+（b-c）²+（a-c）²】/2
=（100+25+225）/2
=175

Answer 3

a-b=10,b-c=5,所以a-c=15
a2+b2+c2-ab-bc-ca
=a(a-b)+b(b-c)+c(c-a)
=10a+5b-15c
=10(a-c)+5(b-c)
=10*15+5*5
=175

Answer 4

因为 a-b=10 b-c=5，所以 a-c=a-b+b-c=15
a²+b²+c²-ab-bc-ca=[(a-b)²+(b-c)²+(a-c)²]/2=125