Question

已知三个不相等的实数a,b,c满足a+1/b=b+1/c=c+1/a=t求t的值

Answer 1

=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