25.已知函数f(x)=lnx−axf(x)=lnx-axf(x)=lnx−ax有两个零点.(1)求aaa的取值范围;(2)设x1x_{1}x1,x2x_{2}x2是f(x)f(x)f(x)的两个零点,证明:a(x1+x2)>2a(x_{1}+x_{2})>2a(x1+x2)>2.