(1)∵f'(x)=(lnx-1)+(x+1)/x=lnx+1/x∴g(x)=lnx-x(x＞0)∴g'(x)=1/x-1令g'(x)=o → x=1∵g(x)在（0,1）上递减,在（1,+∞）上递增∴g(x)max=g(1)=-1(2)(x-1）[f(x)+2]=(lnx-1)(x-1)²≥m∵lnx-1在(0,+∞）上递增,...