计算下面的分式题.急,我是初一的,1、(y-z)(z-x)/[(x+z-2y)(x+y-2z)]+(z-y)(x-z)/[(x+y-2z)(y+z-2x)+(x-z)(y-z)]/[(y+z-2x)(x+z-2y)]2、若a/b=b/c=c/d=d/a,则(a-b+c-d)/(a+b-c+d)的值.3、若x/(x^2-mx+1)=1,求x^3/(x^6-m^3x^3+1)的值.4、如果

计算下面的分式题.急,我是初一的,1、(y-z)(z-x)/[(x+z-2y)(x+y-2z)]+(z-y)(x-z)/[(x+y-2z)(y+z-2x)+(x-z)(y-z)]/[(y+z-2x)(x+z-2y)]2、若a/b=b/c=c/d=d/a,则(a-b+c-d)/(a+b-c+d)的值.3、若x/(x^2-mx+1)=1,求x^3/(x^6-m^3x^3+1)的值.4、如果
计算下面的分式题.急,我是初一的,
1、(y-z)(z-x)/[(x+z-2y)(x+y-2z)]+(z-y)(x-z)/[(x+y-2z)(y+z-2x)+(x-z)(y-z)]/[(y+z-2x)(x+z-2y)]
2、若a/b=b/c=c/d=d/a,则(a-b+c-d)/(a+b-c+d)的值.
3、若x/(x^2-mx+1)=1,求x^3/(x^6-m^3x^3+1)的值.
4、如果a+1/b=1,b+2/c=1,求c+2/a的值.
5、已知xyz=1,x+y+z=2,x^2+y^2+z^2=16,求代数式1/(xy+2z)+1/(yz+2x)+1/(xz+2y)的值.

计算下面的分式题.急,我是初一的,1、(y-z)(z-x)/[(x+z-2y)(x+y-2z)]+(z-y)(x-z)/[(x+y-2z)(y+z-2x)+(x-z)(y-z)]/[(y+z-2x)(x+z-2y)]2、若a/b=b/c=c/d=d/a,则(a-b+c-d)/(a+b-c+d)的值.3、若x/(x^2-mx+1)=1,求x^3/(x^6-m^3x^3+1)的值.4、如果
1、设x-y=a y-z=b z-x=c 则a+b+c=0
(y-z)(z-x)/[(x+z-2y)(x+y-2z)]+(z-y)(x-z)/[(x+y-2z)(y+z-2x)+(z-x)(y-z)]/[(y+z-2x)(x+z-2y)]
=bc/[(a-b)(b-c)]+bc/[(b-c)(c-a)]+bc/[(c-a)(a-b)]
=[bc(c-a)+bc(a-b)+bc(b-c)]/(a-b)(b-c)(c-a)
=0

1.x-y=a
y-z=b
z-x=c
a+b+c=0
∑(z-y)(x-y)(x+z-2y)
=∑(z-y)(x-y)(x-y)+∑(z-y)(x-y)(z-y)
=-∑aab+∑bba
=(a-b)(b-c)(c-a)
1/(x+y-2z)(y+z-2x)(x+z-2y)
=1/(b-c)(c-a)(a-b)
所求...

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1.x-y=a
y-z=b
z-x=c
a+b+c=0
∑(z-y)(x-y)(x+z-2y)
=∑(z-y)(x-y)(x-y)+∑(z-y)(x-y)(z-y)
=-∑aab+∑bba
=(a-b)(b-c)(c-a)
1/(x+y-2z)(y+z-2x)(x+z-2y)
=1/(b-c)(c-a)(a-b)
所求＝1
2.a／b＝b／c＝c／d＝d／a=k
则：a=bk,b=ck,c=dk,d=ak
a=bk=ck^2=dk^3=ak^4
所以k^4=1
如果a,b,c,d都是实数，那么k=+-1
所以得
a=b=c=d
或
a=-b=c=-d
代入要求的式子，得：
(a-b+c-d)/(a+b-c+d)=0或-2
3.设x/(x^2-mx+1)=1，其中m为常数，求x ^3/(x^6-m^3x^3+1)的值详细过程

x/(x^2-mx+1)=1，即(x^2-mx+1)/x=1，即 x + 1/x = 1+m
设x^3/(x^6-m^3x^3+1)的倒数为y
y = (x^6-m^3x^3+1)/x^3
= x^3 + 1/x^3 - m^3
= (x + 1/x)(x^2 + 1/x^2 - 1) - m^3
= (x + 1/x)[(x + 1/x)^2 - 3] - m^3
= (1+m)[(1+m)^2-3] - m^3
= 3m^2 - 2
所以 x^3/(x^6-m^3x^3+1) = 1/y = 1/(3m^2 - 2)
4.由b+2/c=1得
b=1-2/c
代入a+1/b=1得
a+1/（(c-2)/c）=1
a+1+2/(c-2)=1
a=2/(2-c)
代入c+2/a得
c+2/a=c+2/(2/(2-c))=c+2-c=2
5.xy+2z=xy+4-2x-2y=(x-2)(y-2).
yz+2x=(y-2)(z-2),zx+2y=(z-2)(x-2).
4=(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=16+2(xy+yz+zx),
xy+yz+zx=-6.
(x-2)(y-2)(z-2)=xyz-2(xy+yz+zx)+4(x+y+z)-8=13.
原式=[(x-2)+(y-2)+(z-2)]/(x-2)(y-2)(z-2)
=-4/13.

收起

e 我也是初一的，但不会。