已知x²+y²+z²-2x+4y-6z+14=0,则(x-y-z)^2002=
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已知x²+y²+z²-2x+4y-6z+14=0,则(x-y-z)^2002=
已知x²+y²+z²-2x+4y-6z+14=0,则(x-y-z)^2002=
已知x²+y²+z²-2x+4y-6z+14=0,则(x-y-z)^2002=
x²+y²+z²-2x+4y-6z+14=0
x²+y²+z²-2x+4y-6z+1+4+9=0
即x²-2x+1+y²+4y+4+z²-6z+9=0
即(x-1)²+(y+2)²+(z-3)²=0
平方非负
它们的和为0,则只能都是0
∴ x=1,y=-2,z=3
即 (x-y-z)^2002=0^2002=0
x²+y²+z²-2x+4y-6z+14=0
x²-2x+1+y²+4y+4+z²-6z+9=0
(x-1)²+(y+2)²+(z-3)²=0
x-1=0
y+2=0
z-3=0
∴x=1
y=-2
z=3
∴(x-y-z)^2012
=(1+2-3)^2012
=0