已知A+B+C=0 1/(A+1)+1/(B+2)+1/(C+3)=0 求<1/(A+1)>的平方+<1/(B+2)>的平方+<1/(C+3)>的平方=?

来源：百度知道 编辑：UC知道 时间：2024/05/22 20:43:17
"/"是除号 < >是中括号 ( )是小括号

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

36

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

这里用到公式：(x+y+z)^2=(x^2+y^2+z^2)+2(xy+xz+yz)

即：x^2+y^2+z^2=(x+y+z)^2-2(xy+xz+yz)