在af(x)+bf(x/(x-1))=e^x中,x≠1,且x/(x-1) ≠1设t=x/(x-1),得x=t/(t-1),∴af(t)+bf(t/(t-1))=e^t ①af(t/(t-1))+bf(t)=e^[ t/(t-1)] ②①×a-②×b,得(a²-b²)f(t)=a×e^t-b×e^[ t/(t-1)]∵|a|≠|b|,∴a&...