LNM是不是已经属于晚期了
...1)/(x+1)],设m,n属于正有理数,求证:[(m-n)/(lnm-lnn)]<(m+n)/2...
(一)函数f(x)=㏑x-[2(x-1)/(x+1)].(x≥1)求导得:f′(x)=(1/x)-[4/(x+1)²]=(x-1)²/[x(x+1)²]≥0.(x≥1).∴函数f(x)在[1,+∞)上递增。∴当x>1时,f(x)>f(1)=0.即当x>1时,有f(x)>0.即当x>1时,有(㏑x)/2>(x-1)/还有呢?
(一)函数f(x)=㏑x-[2(x-1)/(x+1)].(x≥1)求导得:f′(x)=(1/x)-[4/(x+1)²]=(x-1)²/[x(x+1)²]≥0.(x≥1).∴函数f(x)在[1,+∞)上递增。∴当x>1时,f(x)>f(1)=0.即当x>1时,有f(x)>0.即当x>1时,有(㏑x)/2>(x-1)/还有呢?