已知函数f(x)=(ax²-x)lnx-(1/2)ax²+x,当a=0时,求f(x)在(e,f(e))处切线方程
当a=0时f(x)=-xlnx+x；f(e)=-elne+e=-e+e=0
f '(x)=-lnx,f '(e)=-1,故f(x)在(e,0)处的切线方程为y=-(x-e)=-x+e.