经判断,三个数都不为0
设(a+b)/(a-b)=(b+c)/2(b-c)=(c+a)/3(c-a)=t
得到(t-1)a=(t+1)b
(2t-1)b=(2t+1)c
(3t-1)c=(2t+1)a
三个式子相乘,得打
(t-1)(2t-1)(3t-1)=(t+1)(2t+1)(3t+1)
解得 11t^2+1=0
因为b=(t-1)a/(t+1)
c=(2t+1)a/(3t-1)
所以8a+9b+5c=8a+9(t-1)a/(t+1)+5(2t+1)a/(3t-1)
=6(11t^2+1)a/[(t+1)(3t-1)]
=0