Question

已知a-b=b-c=3/5,a^2+b^2+c^2=1则ab+bc+ca的值等于_

Answer 1

a-b=3/5
b-c=3/5
a-c=6/5
(a-b)²+(b-c)²+(c-a)²=9/25+9/25+36/25
a²-2ab+b²+b²-2bc+c²+c²-2ca+c²=54/25
2(a²+b²+c²)-2(ab+bc+ca)=54/25
2*1-2(ab+bc+ca)=54/25
ab+bc+ca=(2-54/25)/2=-2/25

Answer 2

a-b+b-c=6/5
a-c=6/5

(a-b)²+(b-c)²+(c-a)²=a²-2ab+b²+b²-2bc+c²+c²-2ac+a²
=2a²+2b²+2c²-2ab-2bc-2ac
=2-2(ab+bc+ac)
=(3/5)²+(3/5)²+(6/5)²
=54/25
∴ab+bc+ac=-2/25