问题补充:
若x+17=y+7=z+6,求(x-y)^2+(y-z)^2+(z-x)^2的值
答案:
∵X+17=Y+7
∴X-Y=-10
∵Y+7=Z+6
∴Y-Z=-1
∵X+17=Z+6
∴Z-X=11
∴(x-y)^2+(y-z)^2+(z-x)^2=(-10)^2+(-1)^2+11^2=222
======以下答案可供参考======
供参考答案1:
222 (x-y)^2=100 y-z)^2=1 +(z-x)^2=121 y+7=z+6有y-z=-1
供参考答案2:
由题目可知:x+17=y+7=z+6=0
∴x=-17,y=-7,z=-6
∴原式=[(-17)-(-7)]^2+[(-7)-(-6)]^2+[(-6)-(-17)]^2
=100+1+121
=222