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
问题描述:
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
答
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