Question

已知a-b=b-c=3/5,a^2+b^2+c^2=1,则ab+bc+ac的值如何计算？谢谢

Answer 1

解：∵ a-b=3/5
∴ a=3/5+b
∵ b-c=3/5
∴ b=3/5+c
∴ a-c=6/5
∴ c=a-6/5
ab+bc+ac=(3/5+b)b+(3/5+c)c+a(a-6/5)
= a^2 + b^2 + c^2-6/5a+3/5b+3/5c
=1-3/5a+3/5b-3/5a+3/5c
=1-3/5(a-b)-3/5(a-c)
=1-3/5*3/5-3/5*6/5
=-2/25

Answer 2

a-c=6/5,a=c+6/5,b=c+3/5,代入a^2+b^2+c^2=1得3c^2+18/5*c+45/25=1,
ab+bc+ac=( c+6/5)( c+3/5)+( c+3/5)c+( c+6/5)c
=3c^2+18/5*c+18/25
=1-27/25
=-2/25

Answer 3

1. b-c= 3/5; -> b = 3/5 + c, c= b-3/5, c- b = -3/5;
2. a-b=3/5; -> a = 3/5 +b, b = a - 3/5; -> a = 3/5 + (3/5+c); -> a =6/5 + c, a-c = 6/5;
3.
ab + bc + ac = a(a-3/5) + b(b-3/5) + (6/5+c)c = a^2 - a3/5 +b^2 - b*3/5 +c^2 + c*6/5
= (a^2 + b^2 + c^2) - a*3/5 - b*3/5 + c*6/5
= 1 - a*3/5 - b*3/5 + c*6/5
= 1 + 3/5 (c-a) + 3/5(c-b)
= 1 + 3/5(-6/5) + 3/5(-3/5)
= 1 - 18/25 - 9/25
= -2/25

Answer 4

b-c=3/5 b=3/5+c a-b=a-3/5-c=3/5 a-c=6/5
(a-b)^2+(b-c)^2+(a-c)^2=2(a^2+b^2+c^2)-2(ab+bc+ac)将a-b=b-c=3/5,a^2+b^2+c^2=1，a-c=6/5代入﹙3／5﹚²＋﹙3／5﹚²＋﹙6／5﹚²=2×1-2(ab+bc+ac)
ab+bc+ac=-2／25

Answer 5

这个题只需将那两个式子都平方，然后相加，再减去下面三个数的平方的和。就可以了。哪里用得着代换！！题中还有个隐藏条件a-c=2b

Answer 6

a=b+3/5 c=b-3/5 a^2+b^2+c^2=(3/5+b)^2+(3/5-b)^2+b^2=1 3b^2=7/25 ab+bc+ac=3b^2-9/25=-2/25