9.已知函数f(x)=x−1−alnxf(x)=x-1-alnxf(x)=x−1−alnx,g(x)=xex−1g(x)=\frac{x}{{e}^{x-1}}g(x)=ex−1x.(1)讨论函数f(x)f(x)f(x)的单调性;(2)若a<0a<0a<0,且对任意x1x_{1}x1,x2∈[3x_{2}\in [3x2∈[3,4]4]4](其中x1<x2)x_{1}<x_{2})x1<x2)都有f(x1)−f(x2)g(x1)−g(x2)>−1g(x1)g(x2)\frac{f(x_1)-f(x_2)}{g(x_1)-g(x_2)}>-\frac{1}{g(x_1)g(x_2)}g(x1)−g(x2)f(x1)−f(x2)>−g(x1)g(x2)1,求实数aaa的最小值.