1998(z-y)+1999(y-z)+2000(z-x)=0 1998²(z-y)+1999²(y
4个回答

1998(x-y)+1999(y-z)+2000(z-x)=0 (1)

19982(x-y)+19992(y-z)+20002(z-x)=0 (2)

(1)展开化简后,得

1998x-1998y+1999y-1999z+2000z-2000x=0

-2x+y+z=0 (3)

(2)展开化简后,得

1998^2 x-1998^2 y+1999^2 y-1999^2 z+2000^2 z-2000^2 x=0

(1998^2-2000^2)x+(1999^2-1998^2)y+(2000^2-1999^2)z=0 利用平方差公式,得

-3998*2x+3997y+3999z=0 (4)

(3)两边乘3998-(4),消去x,得

3998y+3998z-(3997y+3999z)=0

z-y=0