已知x-y-z=0,y-z=0,且xyz不等于0,求1998x^2+1999y^2-2000z^2/1998x^2...

发布网友 发布时间：2024-10-23 23:52

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热心网友 时间：2024-10-27 14:14

x-y-z=0
x=y+z
y-z=0
y=z
所以x=y+y=2y
y=z=x/2
y^2=z^2=x^2/4

(1998x^2+1999y^2-2000z^2)/(1998x^2-1999y^2+2000z^2)
=(1998x^2+1999*x^2/4-2000*x^2/4)/(1998x^2-1999*x^2/4+2000*x^2/4)
=(1998+1999/4-2000/4)/(1998-1999/4+2000/4)
=(1998-1/4)/(1998+1/4)
=(4*1998-1)/(4*1998+1)
=7991/7993

1/x+1/y=1/2
(x+y)/xy=1/2
xy=2(x+y)

所以(3x-5xy+3y)/(-x+3xy-y)
=[3(x+y)-5*2(x+y)]/[-(x+y)+3*2(x+y)]
=-7(x+y)/[5(x+y)]
=-7/5

热心网友 时间：2024-10-27 14:07

x-y-z=0,y-z=0
所以：
x=y+z
y=z
x=y+z=2z
带入：
1998x^2+1999y^2-2000z^2/1998x^2-1999y^2+2000z^2
=1998*4z^2+1999z^2-2000z^2/1998*4z^2-1999z^2+2000z^2
=(1998*4+1999-2000)/(1998*4-1999+2000)
=7991/7993

1/x+1/y=5
x+y/xy(通分)=5
∴x+y=5xy
将上式带入原式，并整理：
3x-5xy+3y/x+xy+y
=[3(5xy)-5xy]/[5xy+xy ]
=10xy/6xy
=5/3