已知a²+b²+c²-ab-bc-ac=0,且a+2b+3c=12,则﹙a+b+c﹚²=______
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![已知a²+b²+c²-ab-bc-ac=0,且a+2b+3c=12,则﹙a+b+c﹚²=______](/uploads/image/z/2803114-10-4.jpg?t=%E5%B7%B2%E7%9F%A5a%26%23178%3B%2Bb%26%23178%3B%2Bc%26%23178%3B-ab-bc-ac%3D0%2C%E4%B8%94a%EF%BC%8B2b%EF%BC%8B3c%EF%BC%9D12%2C%E5%88%99%EF%B9%99a%EF%BC%8Bb%EF%BC%8Bc%EF%B9%9A%26%23178%3B%EF%BC%9D______)
已知a²+b²+c²-ab-bc-ac=0,且a+2b+3c=12,则﹙a+b+c﹚²=______
已知a²+b²+c²-ab-bc-ac=0,且a+2b+3c=12,则﹙a+b+c﹚²=______
已知a²+b²+c²-ab-bc-ac=0,且a+2b+3c=12,则﹙a+b+c﹚²=______
a²+b²+c²-ab-bc-ac=0
2a²+2b²+2c²-2ab-2bc-2ac=0
(a-b)²+(a-c)²+(b-c)²=0
a-b=0,a-c=0,b-c=0
a=b=c
∵a+2b+3c=12
∴a=b=c=2
﹙a+b+c﹚²
=(2+2+2)²
=36