A simple ecuation

Call $t = x^{-1}$ .

The equation became

$t^3 -10t^2 +9t = 0$

that is

$t(t-1)(t-9)=0$

whose roots are $0, 1$ and $9$ .

Turning back in $x$ we find $x=1$ and $x = \dfrac{1}{9}$ (the $t=0$ doesn't lead to an acceptable $x$ solution) the least of which is $\dfrac{1}{9}$ .