f(x)=xlnx/(x+1) (1)若任取x∈[1,+∞﹚,f(x)≤m(x-1)恒成立,求m范围(2)求证ln(2n+1)^1/4<i/(4i^2-1)第二问证ln(2n+1)^1/4<{i/(4i^2-1)(i从1开始取,前n项和∑)}

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f(x)=xlnx/(x+1) (1)若任取x∈[1,+∞﹚,f(x)≤m(x-1)恒成立,求m范围(2)求证ln(2n+1)^1/4<i/(4i^2-1)第二问证ln(2n+1)^1/4<{i/(4i^2-1)(i从1开始取,前n项和∑)}

f(x)=xlnx/(x+1) (1)若任取x∈[1,+∞﹚,f(x)≤m(x-1)恒成立,求m范围(2)求证ln(2n+1)^1/4<i/(4i^2-1)第二问证ln(2n+1)^1/4<{i/(4i^2-1)(i从1开始取,前n项和∑)}
f(x)=xlnx/(x+1) (1)若任取x∈[1,+∞﹚,f(x)≤m(x-1)恒成立,求m范围
(2)求证ln(2n+1)^1/4<i/(4i^2-1)
第二问证ln(2n+1)^1/4<{i/(4i^2-1)(i从1开始取,前n项和∑)}

f(x)=xlnx/(x+1) (1)若任取x∈[1,+∞﹚,f(x)≤m(x-1)恒成立,求m范围(2)求证ln(2n+1)^1/4<i/(4i^2-1)第二问证ln(2n+1)^1/4<{i/(4i^2-1)(i从1开始取,前n项和∑)}
1、即,使xlnx/(x^2-1)≤m在x∈[1,+∞﹚上恒成立
令g(x)=xlnx/(x^2-1)
g'(x)=[(lnx+1)(x^2-1)-2x^2lnx]/(x^2-1)^2=[-x^2lnx+x^2-lnx-1]/(x^2-1)^2
令h(x)=-x^2lnx+x^2-lnx-1
h'(x)=-2xlnx+x-1/x
h''(x)=-2lnx+1/(x^2)-1
h'''(x)=-2/x-2/x^3=-(2/x)*(1+1/x^2)