已知实数a,b,c满足a2-2b=-2,b2+6c=7,c2-8a=-31,则a+b+c的值等于 ________.
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解析分析:首先将三个式子左边与右边分别相加,即可得:a2-2b+b2+6c+c2-8a+26=0,再将其配方,得(a-4)2+(b-1)2+(c+3)2=0,由非负数的和为0,每个为0,即可求得结果.
解答:∵a2-2b=-2,b2+6c=7,c2-8a=-31,∴a2-2b+b2+6c+c2-8a+26=0,∴(a-4)2+(b-1)2+(c+3)2=0,∴a-4=0,b-1=0,c+3=0,∴a=4,b=1,c=-3,∴a+b+c=2.故