若x+y+z=0,求x2+y2+z2-xy-xz-zy的值

如题所述

第1个回答  2012-02-21
由x+y+z=0得(x+y+z)2=0;
于是可得x2+y2+z2+2(xy+yz+zx)=0;
所以有xy+yz+zx=-(x2+y2+z2)/2;
则有:x2+y2+z2-xy-yz-zx=x2+y2+z2-(xy+yz+zx)=x2+y2+z2-[-(x2+y2+z2)/2]=x2+y2+z2+(x2+y2+z2)/2=3(x2+y2+z2)/2

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