Kreska ułamkowa oznacza dzielenie. Nie można dzielić przez 0, więc musimy zadbać o to, aby wyrażenia w mianownikach nie przyjmowały wartości 0. 

 

 

$a)$  

$\{\begin{array}{l}x\ne 0\\ x-2\ne 0\mid +2\end{array}$  

$\{\begin{array}{l}x\ne 0\\ x\ne 2\end{array}$  

$D=\mathbb{R}\backslash \left\{0;2\right\}$  

 

$b)$  

$\{\begin{array}{l}x+3\ne 0\mid -3\\ 5-x\ne 0\mid +x\end{array}$  

$\{\begin{array}{l}x\ne -3\\ 5\ne x\end{array}$  

$D=\mathbb{R}\backslash \left\{-3;5\right\}$  

 

 

$c)$  

$\{\begin{array}{l}4x+3\ne 0\mid -3\\ 6-x\ne 0\mid -6\end{array}$  

$\{\begin{array}{l}4x\ne -3\mid :4\\ -x\ne -6\mid \cdot \left(-1\right)\end{array}$  

$\{\begin{array}{l}x\ne -\frac{3}{4}\\ x\ne 6\end{array}$  

$D=\mathbb{R}\backslash \left\{-\frac{3}{4};6\right\}$  

 

 

$d)$  

$\{\begin{array}{l}8x-6\ne 0\mid +6\\ x+1\ne 0\mid -1\end{array}$  

$\{\begin{array}{l}8x\ne 6\mid :8\\ x\ne -1\end{array}$  

$\{\begin{array}{l}x\ne \frac{3}{4}\\ x\ne -1\end{array}$  

$D=\mathbb{R}\backslash \left\{-1;\frac{3}{4}\right\}$  

 

 

$e)$  

$x\left(x+6\right)\ne 0$  

$\{\begin{array}{l}x\ne 0\\ x+6\ne 0\mid -6\end{array}$  

$\{\begin{array}{l}x\ne 0\\ x\ne -6\end{array}$  

$D=\mathbb{R}\backslash \left\{-6,0\right\}$  

 

 

$f)$  

$\left(x-1\right)\left(x+3\right)\ne 0$  

$\{\begin{array}{l}x-1\ne 0\mid +1\\ x+3\ne 0\mid -3\end{array}$   

$\{\begin{array}{l}x\ne 1\\ x\ne -3\end{array}$    

$D=\mathbb{R}\backslash \left\{-3,1\right\}$