Question

(a+1)²+(b+2)的平方+(c+3)

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

Answer 1

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