数学导数系数取值问题求解已知函数f(x)=(lnx)/(x+1)+1/x,如果x大于0x≠1时,f(x)大于lnx/(x-1)+k/x,求k的取值范围.要过程,分不是问题,谢了.

李漪回答：

　　令y=(lnx)/(x+1)+1/x-lnx/(x-1)-k/x→当x＞0,且x≠1时,xlnx/(x-1)＜(1-k)/2.令z=xlnx/(x-1)→z'=[x-1-(x+1)lnx]/(x-1)令w=x-1-(x+1)lnx→w'=x-1/x-2xlnx→w''=1/x-2lnx-1→w'''=-2(x+1)/x易知：当x＞0时,w'''＜0→w''单调递减.当x=1时,w''=0→当0＜x＜1时,w''＞0→w'单调递增；当x＞1时,w''＜0→w'单调递减.当x=1时,w'=0→当x＞0,且x≠1时,w'＜0→w单调递减.当x=1时,w=0→当0＜x＜1时,w＞0→z单调递增；当x＞1时,w＜0→z单调递减.易得：lim(x→1)xlnx/(x-1)=1/2→当x＞0,且x≠1时,xlnx/(x-1)＜1/2由xlnx/(x-1)＜(1-k)/2,xlnx/(x-1)＜1/2→k≤0故：k的取值范围是k≤0