Question

已知a=2005x+2006,b=2005x+2005,c=2005x+2007,求代数式a²+b²+c²-ab-bc-ac

Answer 1

a-b=2005x+2006-2005x-2005=1
同理
b-c=-2
c-a=1
所以原式=(2a²+2b²+2c²-2ab-2bc-2ac)/2
=[(a²-2ab+b²)+(b²-2bc+c²)+(c²-2ac+a²)]/2
=[(a-b)²+(b-c)²+(c-a)²]/2
=(1+4+1)/2
=3

Answer 2

a²+b²+c²-ab-bc-ac
=½﹙2a²+2b²+2c²-2ab-2ac-2bc)
=½[(a²-2ab+b²)+(a²-2ac+c²)+(b²-2bc+c²﹚]
=½[﹙a-b)²+(a-c)²+(b-c)²]
=½×﹙1²+1²+2²)
=3

Answer 3

a-b=1，b-c=-2，a-c=-1
所以原式=1/2×(2a²+2b²+2c²-2ab-2bc-2ac)
=1/2×[(a²-2ab+b²)+(b²-2bc+c²)+(a²-2ac+c²)]
=1/2×[(a-b)²+(b-c)²+(a-c)²]
=1/2×(1+4+1)
=3

Answer 4

a²+b²+c²-ab-bc-ac=a(a-b)+b(b-c)+c(c-a)=(2005x+2006)(2005x+2006-2005x-2005)+(2005x+2005)(2005x+2005-2005x-2007)+(2005x+2007)(2005x+2007-2005x-2006)
=2005x+2006+2005x+2007-2(2005x+2005)=3