Question

已知互不相等的实数a、b、c满足a+1/b=b+1/c=c+1/a=t,则t=?

Answer 1

因为已知互不相等的实数a、b、c满足a+1/b=b+1/c=c+1/a=t
令b=1,
则a+1/b=b+1/c
a+1=1+1/c
a=1/c
b+1/c=c+1/a
1+1/c=c+1/(1/c)=2c
1+1/c=2c
2c^2-c-1=0
(2c+1)(c-1)=0
得c=1或c=-1/2
因为a、b、c互不相等，所以c=-1/2
a=1/c=-2
所以t=a+1/b=-2+1=-1

Answer 2

解：
b=1/(t-a)
c=(t-a)/(t^2-at-1)
又c+1/a=t,c= t - 1/a = (t-a)/(t^2-at-1)
展开得 at^3-(a^2+1)t^2 - at + a^2 + 1 = 0
两边除以a,得
t^3 - ( a + 1/a )t^2 - t + ( a + 1/a ) = 0
设x=a + 1/a,得
x^3 - xt^2 - t+ x=0
分解因式
( t - x) ( t+ 1 ) ( t - 1 ) =0.
若t=x,则a+1/b = t = a+1/a,得b=a,与题意不符.
所以 t=1 或 t=-1