普路通有前景吗:初二代数 help!!!!!!!!

来源：百度文库编辑：查人人中国名人网时间：2024/04/26 12:59:47

已知a-b=3 b-c=2
求a^2+b^2+c^2-ab-bc-ac

a-b=3
b-c=2
=>a-b+b-c=a-c=5
a^2+b^2+c^2-ab-bc-ac
=[(a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)]/2
=[(a-b)^2+(b-c)^2+(a-c)^2]/2
=(3^2+2^2+5^2)/2
=(9+4+25)/2
=19

a^2+b^2+c^2-ab-bc-ac
=2(a^2+b^2+c^2-ab-bc-ac)/2
=(2a^2+2b^2+2c^2-2ab-2bc-2ac)/2
=(a^2-2ab+b^2+a^2-2ac+c^2+b^-2bc+c^2)/2
=(a-b)^2(a-c)^2(b-c)^2/2
因为a-b=3 b-c=2
所以a-c=5
把这三个式子代入
(a-b)^2(a-c)^2(b-c)^2/2
=3^2 2^2 5^2 /2
=450

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