$f(x) = x+\frac{1}{x^x}= x+x^{-x} \Rightarrow f' =1 -x^x(\ln x+ 1)$
Ta CM $f' > 0 \forall x\in (0, 1)$
$\Leftrightarrow x^x(\ln x+1) < 1$
Do $x\in (0,1) \Rightarrow 0<x^x<1, \ln x+ 1<1$ , suy ra $f(x)' > 0$
$\Rightarrow f(x)< f(1)= 2\blacksquare \blacksquare \blacksquare $
- Ispectorgadget yêu thích