问题描述: 若a+b+c=0,1a+1+1b+2+1c+3=0 1个回答 分类:数学 2014-10-07 问题解答: 我来补答 ∵a+b+c=0,∴(a+1)+(b+2)+(c+3)=6,两边平方得(a+1)2+(b+2)2+(c+3)2+2[(a+1)(b+2)+(a+1)(c+3)+(b+2)(c+3)]=36,又由1a+1+1b+2+1c+3=0去分母,得(b+2)(c+3)+(a+1)(c+3)+(a+1)(b+2)=0,∴(a+1)2+(b+2)2+(c+3)2=36.故答案为:36. 展开全文阅读