a²+b²+c²-ab-ac-bc
=1/2(2a²+2b²+2c²-2ab-2ac-2bc)
=1/2[(a²-2ab+b²)+(a²-2ac+c²)+(b²-2bc+c²)]
=1/2[(a-b)²+(a-c)²+(b-c)²]
=1/2×[(-1)²+(-2)²+(-1)²]
=1/2×(1+1+4)
=1/2×6
=3
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a²+b²+c²-ab-ac-bc
=1/2(2a²+2b²+2c²-2ab-2ac-2bc)
=1/2[(a²-2ab+b²)+(a²-2ac+c²)+(b²-2bc+c²)]
=1/2[(a-b)²+(a-c)²+(b-c)²]
=1/2×[(-1)²+(-2)²+(-1)²]
=1/2×(1+1+4)
=1/2×6
=3
土豆团邵文潮为您答疑解难.
如果本题有什么不明白可以追问,