$\frac{x^2-4}{x^2-x-2}=\frac{\left(x-2\right)\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}$

Dziedzina:

$\left(x-1\right)\left(x+2\right)\ne 0$

$x-1\ne 0\text{i}x+2\ne 0$

$x\ne 1\text{i}x\ne -2$

$D=\mathbb{R}\setminus \left\{-2,1\right\}$

Skracanie:

$\frac{\left(x-2\right)\cancel{\left(x+2\right)}}{\left(x-1\right)\cancel{\left(x+2\right)}}=\frac{x-2}{x-1}$