已知函数f(x)=log2((x-1)/(x+1)),g(x)=2ax+1-a,又h(x)=f(x)+g(x)a=1时a=1时,求证h(x)在x属于(1,+∞)上单调递增,并证明函数h(x)有两个零点;a=1时,h(x)=f(x)+g(x)=log(2)[(x-1)/(x+1)]+2x=log(2)(x-1)-log(2)(x+1)+2x求导得x∈(1,

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