Question

x-3=y-2=z-1，求x²+y²+z²-xy-yz-zx的值。

Answer 1

x-3=y-2=z-1
x²+y²+z²-xy-yz-zx
=1/2(2x²+2y²+2z²-2xy-2yz-2zx)
=1/2[(x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2zx+x^2)]
=1/2[(x-y)^2+(y-z)^2+(z-x)^2] x-3=y-2z-1 x-y=1 y-z=1 z-x=-2
=1/2[1+1+4]=3

追问

为什么要提½

追答

要凑成x-y y-z z-x的完全平方式

追问

x-3=y-2=z-1怎么得出x-y=1 y-z=1 z-x=2

追答

x-3=y-2 x-y=1
y-2=z-1 y-z=1
x-3=z-1 z-x=-2

Answer 2

x-3=y-2=z-1得出x-y=1,x-z=2,y-z=1.
2（x²+y²+z²-xy-yz-zx）/2
=[(x-y)²+(y-z)²+(x-z)²]/2
=[1+1+4]/2
=3

Answer 3

X=Y+1,
Z=Y-1
x²+y²+z²-xy-yz-zx
=(Y+1)²+y²+(Y-1)²-y(Y+1)-y(Y-1)-(Y-1)(Y+1)
=3