Question

求教，已知1999(x-y)+2000(y-z)-2001(x-z）=0

已知1999(x-y)+2000(y-z)-2001(x-z）=0

1999 ^2(x-y）+2002＾2（y-z）＋2001^2(z-x)=2000
求z-y的值

Answer 1

• 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

Answer 2

检查一下题目有错误没?