作业帮已知函数f(x)=ln(x+1)/(x-1)(Ⅰ)求函数的定义域.并证明f(x)=ln(x+1)/(x-1)在定义域上是奇函数（Ⅱ）若x属于[2,6]f(x)=ln(x+1)/(x-1)＞ln(m)/(x-1)(x-7)恒成立,求实数m的取值范围
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![已知函数f(x)=ln(x+1)/(x-1)(Ⅰ)求函数的定义域.并证明f(x)=ln(x+1)/(x-1)在定义域上是奇函数（Ⅱ）若x属于[2,6]f(x)=ln(x+1)/(x-1)＞ln(m)/(x-1)(x-7)恒成立,求实数m的取值范围](/uploads/image/z/249431-23-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dln%28x%2B1%29%2F%28x-1%29%28%E2%85%A0%29%E6%B1%82%E5%87%BD%E6%95%B0%E7%9A%84%E5%AE%9A%E4%B9%89%E5%9F%9F.%E5%B9%B6%E8%AF%81%E6%98%8Ef%28x%29%3Dln%28x%2B1%29%2F%28x-1%29%E5%9C%A8%E5%AE%9A%E4%B9%89%E5%9F%9F%E4%B8%8A%E6%98%AF%E5%A5%87%E5%87%BD%E6%95%B0%EF%BC%88%E2%85%A1%EF%BC%89%E8%8B%A5x%E5%B1%9E%E4%BA%8E%5B2%2C6%5Df%28x%29%3Dln%28x%2B1%29%2F%28x-1%29%EF%BC%9Eln%28m%29%2F%28x-1%29%28x-7%29%E6%81%92%E6%88%90%E7%AB%8B%2C%E6%B1%82%E5%AE%9E%E6%95%B0m%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
已知函数f(x)=ln(x+1)/(x-1)(Ⅰ)求函数的定义域.并证明f(x)=ln(x+1)/(x-1)在定义域上是奇函数（Ⅱ）若x属于[2,6]f(x)=ln(x+1)/(x-1)＞ln(m)/(x-1)(x-7)恒成立,求实数m的取值范围
已知函数f(x)=ln(x+1)/(x-1)(Ⅰ)求函数的定义域.并证明f(x)=ln(x+1)/(x-1)在定义域上是奇函数
（Ⅱ）若x属于[2,6]f(x)=ln(x+1)/(x-1)＞ln(m)/(x-1)(x-7)恒成立,求实数m的取值范围
已知函数f(x)=ln(x+1)/(x-1)(Ⅰ)求函数的定义域.并证明f(x)=ln(x+1)/(x-1)在定义域上是奇函数（Ⅱ）若x属于[2,6]f(x)=ln(x+1)/(x-1)＞ln(m)/(x-1)(x-7)恒成立,求实数m的取值范围
已知函数f(x)=ln[(x+1)/(x-1)],(Ⅰ)求函数的定义域.并证明f(x)=ln[(x+1)/(x-1)]在定义域上是奇函数；（Ⅱ）若x属于[2,6],f(x)=ln[(x+1)/(x-1)]＞ln[m/(x-1)(x-7)]恒成立,求实数m的取值范围
(1).定义域：由(x+1)/(x-1)>0,得定义域为x1,即定义域为(-∞,-1)∪(1,+∞)；
定义域关于原点对称,且f(-x)=ln[(-x+1)/(-x-1)]=ln[(x-1)/(x+1)]=ln[(x+1)/(x-1)]ֿ¹=-ln[(x+1)/(x-1)
=-f(x),故f(x)是奇函数.
(2).y=lnx是增函数,故由ln[(x+1)/(x-1)]＞ln[m/(x-1)(x-7)]得(x+1)/(x-1)>m/(x-1)(x-7)
移项得(x+1)/(x-1)-m/(x-1)(x-7)=[(x+1)(x-7)-m]/(x-1)(x-7)=(x²-6x-7-m)/(x-1)(x-7)>0.(1)
不等式(1)在区间[2,6]上恒成立,此时有x-1>0,x-7
右边除到左边，证明单调递加，最小值大于0
定义域是-1