x>0f'(x)=(lnx-1)/(lnx)²所以当0当x>1时,f'(x)>0,f(x)单调递增单调递减区间为(0,1),单调递增区间为(1,+∞)(2)x>迟返友1时,lnx>0,x-1>0,故只需证明lnx设g(x)=lnx-x+1,x>1g'(x)=1/x-1<0所以g(x)单码槐调递减,g(x)所以lnx故x/lnx>x/(x-1)所以f(x)>x/(x-1)