已知1999(x-y)+2000(y-z)-2001(x-y)=01999²(x-y)+2000²(y-z)+2001²(y-z)=2000,求z-y的值1式的2001(x-y)应改为2001(x-z),二式的2001²(y-z)应改为2001²(z-x)

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已知1999(x-y)+2000(y-z)-2001(x-y)=01999²(x-y)+2000²(y-z)+2001²(y-z)=2000,求z-y的值1式的2001(x-y)应改为2001(x-z),二式的2001²(y-z)应改为2001²(z-x)
已知1999(x-y)+2000(y-z)-2001(x-y)=0
1999²(x-y)+2000²(y-z)+2001²(y-z)=2000,求z-y的值
1式的2001(x-y)应改为2001(x-z),二式的2001²(y-z)应改为2001²(z-x)

已知1999(x-y)+2000(y-z)-2001(x-y)=01999²(x-y)+2000²(y-z)+2001²(y-z)=2000,求z-y的值1式的2001(x-y)应改为2001(x-z),二式的2001²(y-z)应改为2001²(z-x)
已知1999(x-y)+2000(y-z)+2001(x-z)=0
∴1999²(x-y)+2000²(y-z)+2001²(z-x)=2000,求z-y的值
1999(x-y)+2000(y-z)-2001(x-z)=0
1999x-1999y+2000y-2000z-2001x+2001z=0
-2x+y+z=0
∴x=(y+z)/2
1999²(x-y)+2000²(y-z)+2001²(z-x)=2000
1999²[(y+z)/2-y]+2000²(y-z)+2001²[(z-(y+z)/2]=2000
1999²(z-y)/2+2000²(y-z)+2001²(z-y)/2=2000
(z-y)(1999²-2*2000²+2001²)=2000
(z-y)[1999²-2*(1999+1)²+2001²]=2000
(z-y)(1999²-2*1999²-4*1999-2+2001²)=2000
(z-y)(2001²-1999²-4*1999-2)=2000
(z-y)(8000-7996-2)=2000
(z-y)*2=2000
∴z-y=1/1000

已知1999(x-y)+2000(y-z)+2001(x-z)=0
因为1999²(x-y)+2000²(y-z)+2001²(z-x)=2000,求z-y的值

1999(x-y)+2000(y-z)-2001(x-z)=0
1999x-1999y+2000y-2000z-2001x+2001z=0
-2x+y+z=0

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已知1999(x-y)+2000(y-z)+2001(x-z)=0
因为1999²(x-y)+2000²(y-z)+2001²(z-x)=2000,求z-y的值

1999(x-y)+2000(y-z)-2001(x-z)=0
1999x-1999y+2000y-2000z-2001x+2001z=0
-2x+y+z=0
所以x=(y+z)/2
1999²(x-y)+2000²(y-z)+2001²(z-x)=2000
1999²[(y+z)/2-y]+2000²(y-z)+2001²[(z-(y+z)/2]=2000
1999²(z-y)/2+2000²(y-z)+2001²(z-y)/2=2000
(z-y)(1999²-2*2000²+2001²)=2000
(z-y)[1999²-2*(1999+1)²+2001²]=2000
(z-y)(1999²-2*1999²-4*1999-2+2001²)=2000
(z-y)(2001²-1999²-4*1999-2)=2000
(z-y)(8000-7996-2)=2000
(z-y)*2=2000所以
z-y=1/1000

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