(1)化简(a-b)²+(a-c)²+(c-a)² (2)利用上题结论,且a-b=10,b-c=5,求化简a²+b²+c²-ab-bc-ca的值
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![(1)化简(a-b)²+(a-c)²+(c-a)² (2)利用上题结论,且a-b=10,b-c=5,求化简a²+b²+c²-ab-bc-ca的值](/uploads/image/z/5429220-60-0.jpg?t=%EF%BC%881%EF%BC%89%E5%8C%96%E7%AE%80%28a-b%29%26%23178%3B%2B%28a-c%29%26%23178%3B%2B%28c-a%29%26%23178%3B+%EF%BC%882%EF%BC%89%E5%88%A9%E7%94%A8%E4%B8%8A%E9%A2%98%E7%BB%93%E8%AE%BA%2C%E4%B8%94a-b%3D10%2Cb-c%3D5%2C%E6%B1%82%E5%8C%96%E7%AE%80a%26%23178%3B%2Bb%26%23178%3B%2Bc%26%23178%3B-ab-bc-ca%E7%9A%84%E5%80%BC)
(1)化简(a-b)²+(a-c)²+(c-a)² (2)利用上题结论,且a-b=10,b-c=5,求化简a²+b²+c²-ab-bc-ca的值
(1)化简(a-b)²+(a-c)²+(c-a)² (2)利用上题结论,且a-b=10,b-c=5,求化简a²+b²+c²-ab-bc-ca的值
(1)化简(a-b)²+(a-c)²+(c-a)² (2)利用上题结论,且a-b=10,b-c=5,求化简a²+b²+c²-ab-bc-ca的值
(a-b)²+(b-c)²+(c-a)²
=a²-2ab+b²+b²-2bc+c²+c²-2ac+a²
=2a²+2b²+2c²-2ab-2bc-2ac
a-b==10
b-c=5
相加
a-c=15
c-a=-15
所以原式=[(a-b)²+(b-c)²+(c-a)² ]/2
=(100+25+225)/2
=175
(a-b)²+(b-c)²+(c-a)²
=a²-2ab+b²+b²-2bc+c²+c²-2ac+a²
=2a²+2b²+2c²-2ab-2bc-2ac
全×2