(1)、f'(x)=2x-1-1/x (x>0)令f'(x)=0,即2x-1-1/x =0方程两边同乘以x得：2x²-x-1=0(x-1)(2x+1)=0解得：x=1或-1/2,其中x=-1/2不合题意,舍去当00,即f(x)单调递增∴当x=1时,f(x)取得极小值f(1)=1²-1-ln1=0(2)...