若x+17=y+7=z+6,求(x-y)^2+(y-z)^2+(z-x)^2的值

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若x+17=y+7=z+6,求(x-y)^2+(y-z)^2+(z-x)^2的值
若x+17=y+7=z+6,求(x-y)^2+(y-z)^2+(z-x)^2的值

若x+17=y+7=z+6,求(x-y)^2+(y-z)^2+(z-x)^2的值
∵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

222 (x-y)^2=100 y-z)^2=1 +(z-x)^2=121 y+7=z+6有y-z=-1

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

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