Question

若a/b=b/c=c/d=d/a,则(a-b+c-d)/(a+b+d-c)的值为

Answer 1

a/b=b/c=c/d=d/a,则(a-b+c-d)/(a+b-c+d)的值是
设a/b=b/c=c/d=d/a=t

d=at,
c=dt=t*at=at^2
b=ct,
b=at^2*t=at^3
a=bt
a=at^3*t=at^4
t^4=1
所以
t=1或者
t=-1
(a-b+c-d)/(a+b-c+d)=(at^4-at^3+at^2-at)/(at^4+at^3-at^2+at)=(t^4-t^3+t^2-t)/(t^4+t^3-t^2+t)=(t^3-t^2+t-1)/(t^3+t^2-t+1)
当t=1时，原式=（1-1+1-1）/(1+1+-1+1)=0
当t=-1时，原式=[-1-(-1)^2+(-1)-1]/[-1+(-1)^2-(-1)+1]

=-2
综述
：
a/b=b/c=c/d=d/a=1时，(a-b+c-d)/(a+b-c+d)=0

a/b=b/c=c/d=d/a=-1时，(a-b+c-d)/(a+b-c+d)=-2

Answer 2

设a/b=b/c=c/d=d/a=k,则a=b*k,b=c*k,c=d*k,d=a*k

∴a=c*k^2,b=c*k,d=c*k^3
c=c*k^4推出k=±1（不考虑虚数）

(a-b+c-d)/(a+b+d-c)=(c*k^2-c*k+c-c*k^3)/(c*k^2+c*k+c*k^3-c)=(k^2-k+1-k^3)/(k^2+k+k^3-1)
1.若k=1，原式=0
2.若k=-1，原式=-2

Answer 3

易得a=b=c=d