Question

求a平方+b平方+c平方-ab-bc-ac

a=二分之一m+2000，b=二分之一m+2001，c=二分之一m+2002，求a平方+b平方+c平方-ab-bc-ac

Answer 1

a^2+b^2+c^2-ab-bc-ac
=2(a^2+b^2+c^2-ab-bc-ac)/2
=〔(a-b)^2+(a-c)^2+(b-c)^2〕/2
=〔(a-b)^2+(a-c)^2+(b-c)^2〕/2
把a=1/2+2000,b=1/2+2001,c=1/2+2002代入上式化简有
〔(a-b)^2+(a-c)^2+(b-c)^2〕/2
=〔(-1)^2+(-2)^2+(-2)^2〕/2
＝3

Answer 2

解：

a^2+b^2+c^2-ab-bc-ac
=(1/2)(2a^2+2b^2+2c^2-2ab-2bc-2ca)
=(1/2)[(b-a)^2+(c-b)^2+(c-a)^2]
=(1/2)(1^2+1^2+2^2)
=6/2
=3

Answer 3

a^2+b^2+c^2-ab-bc-ca
=(2a^2+2b^2+2c^2-2ab-2bc-2ca)/2
=(a^2+b^2-2ab+a^2+c^2-2ca+b^2+c^2-2bc)/2
=[(a-b)^2+(c-a)^2+(b-c)^2]/2
=[(-1)^2+2^2+(-1)^2]/2
=(1+4+1)/2
=3

Answer 4

3