Question

完全平方公式

a=2010x+2009,b=2010x+2010,c=2010x+2011,求多项式a²+b²+c²-ab-bc-ac的值

Answer 1

2a²+2b²+2c²-2ab-2bc-2ac=（a²-2ab+b²）+（b²-2bc+c²）+（a²-2ac+c²）=（a-b）²+（a-c）²+（b-c）²=1+1+4=6
所以原式等于3
祝学习进步~

Answer 2

解：a²+b²+c²-ab-bc-ac
=(2a²+2b²+2c²-2ab-2bc-2ac)/2
=(a²+b²-2ab+b²+c²-2bc+a²+c²-2ac)/2
=[(a-b)²+(b-c)²+(a-c)²]/2
=[（2010x+2009-2010x-2010)²+(2010x+2010-2010x-2011)²+(2010x+2009-2010x-2011)²]/2
=[(-1)²+(-1)²+(-2)²]/2
=6/2
=3

Answer 3

a²+b²+c²-ab-bc-ac
=1/2[2(a²+b²+c²-ab-bc-ac)]
=1/2(2a²+2b²+2c²-2ab-2bc-2ac)
=1/2[(a²-2ab+b²)+(a²-2ac+c²)+(b²-2bc+c²)]
=1/2[(a-b)²+(a-c)²+(b-c)²]
=1/2{[(2010x+2009)-(2010x+2010)]²+[(2010x+2009)-(2010x+2011)]²+[(2010x+2010)-(2010x+2011)]²}
=1/2(1+4+1)
=3

Answer 4

a²+b²+c²-ab-bc-ac=((a-b)²+(b-c)²+(c-a)²)/2=(1+1+4)/2=3

Answer 5

a2+b2+c2-ab-bc-ac=1/2[（a^2-2ab+b^2）+(a^2-2ac+c^2)+(b^2-2bc+c^2)]
=1/2[(a-b)^2+(a-c)^2+(b-c)^2]=1/2(1+4+1)=3

Answer 6

多项式=(1/2)[(a-b)^2+(b-c)^2+(a-c)^2]=(1/2)(1+1+2^2)=3