x(1+lnx)
问题描述：
x(1+lnx)
答
f(x)= x(1 + lnx) + 1 = x + xlnx + 1
f'(x) = 1 + lnx + x*1/x = 2 + lnx = 0
x = 1/e²
0 x > 1/e²:f'(x) > 0
f(1/e²)为最小值,f(1/e²) = (1/e²)[1 + ln(1/e²)] + 1 = 1 + (1/e²)(1 - 2) = 1 - 1/e² > 0
不等式无解