如果a+b+c=0且(1/a+1)+(1/b+2)+(1/c+3)=0 那麽(a+1)(a+1)+(b+2)(b+2)+(1/c+3)=?

来源：百度知道 编辑：UC知道 时间：2024/05/26 08:20:14
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36

a+1,b+2,c+3 <>0

1/(a+1)+1/(b+2)+1/(c+3) = 0
两边乘 (a+1)(b+2)(c+3)
(b+2)(c+3)+(a+1)(c+3)+(a+1)(b+2) = 0

(a+1)2+(b+2)2+(c+3)2
= (a+1+b+2+c+3)2 - 2[(a+1)(b+2)+(a+1)(c+3)+(b+2)(c+3)]
= (a+b+c+6)2 - 0
= 36

a+b+c=0,所以(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
那么，（a+1)的平方+（b+2)的平方+（c+3)的平方
＝〔(a+1)+(b+2)+(c+3)〕＾2－2〔(b+2)(a+1)+(b+2)(c+3)+(a+1)(c+3)〕
＝36－0
＝36