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a+b+c=0,1/(a+1)+1/(b+2)+1/(c+3)=0,求(a+1)的平方+(b+2)的平方+(c+3)的平

来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/09 13:53:24
a+b+c=0,1/(a+1)+1/(b+2)+1/(c+3)=0,求(a+1)的平方+(b+2)的平方+(c+3)的平方的值
a+b+c=0,(a+1)+(b+2)+(c+3)=6,
1/(a+1)+1/(b+2)+1/(c+3)=0,得(b+2)(a+1)+(b+2)(c+3)+(a+1)(c+3)=0
(a+1)^2+(b+2)^2+(c+3)^2
=〔(a+1)+(b+2)+(c+3)〕^2-2〔(b+2)(a+1)+(b+2)(c+3)+(a+1)(c+3)〕
=36