f'(x)=1-1/x-(1-lnx)/x^2=(x^2-x-1+lnx)/x^2,对x^2-x-1+lnx求导,得2x-1+1/x=(2x^2-x+1)/x,因为2x^2-x+1=2(x-1/4)^2+7/8>0恒成立,所以x^2-x-1+lnx恒增,且当x=1时,x^2-x-1+lnx0,所以用二分法可知x无限逼近1,所以取x=1...X属于（0，+oo)