若x-2008=y-2007=z-2006,求a^2+y^2+x^2-xy-yz-zx的值

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∵x-2008=y-2007=z-2006
∴x-y=1 y-z=1 x-z=2
x²+y²+z²-xy-yz-xz
=(2x²+2y²+2z²-2xy-2yz-2xz)÷2
=[(x-y)²+(y-z)²+(x-z)²]÷2
=(1²+1²+2²)÷2
=6÷2=3======以下答案可供参考======
供参考答案1:
x^2+y^2+z^2-xy-yz-zx
=x^2-xy+y^2-yz+z^2-zx
=x*（x-y）+y*(y-z）+z*（z-x）
x-2008=y-2007=z-2006
则 x-y=1 y-z=1 z-x=-2 即x-z=2
则 x^2+y^2+z^2-xy-yz-zx
=x+y-2z
=x-z+y-z
=2+1=3供参考答案2:
可知x-y=1,y-z=1,z-x=-2
原式=[(x-y)^2+(y-z)^2+(z-x)^2]/2=(1+1+4)/2=3
供参考答案3:
x-2008=y-2007 x-y=1 X=Y+1
y-2007=z-2006 y-z=1 Z=Y-1
x-2008=z-2006 x-z=2
x^2+y^2+Z^2-xy-yz-zx
=(x^2-XY)+(y^2-YZ)+(Z^2-ZX)
=X(X-Y)+Y(Y-Z)-Z(X-Z)
=X+Y-2Z
=Y+1+Y-2(Y-1)
=3