Simplify the expressions below: 1.1.1 3x-1 . (1/3) . (0,125)2x-1 / 27x-x+2

We have $3^{x-1}$, which is already in base 3.
$\left(\frac{1}{3}\right)^{-2} = (3^{-1})^{-2} = 3^{(-1) \times (-2)} = 3^2$.
$0.125 = \frac{125}{1000} = \frac{1}{8} = 2^{-3}$. So, $(0.125)^{2x-1} = (2^{-3})^{2x-1} = 2^{-3(2x-1)} = 2^{-6x+3}$.
$27^{-x+2} = (3^3)^{-x+2} = 3^{3(-x+2)} = 3^{-3x+6}$.

Using the rule $a^m \cdot a^n = a^{m+n}$:

Using the rule $\frac{a^m}{a^n} = a^{m-n}$: