$a)$  

$1:\frac{x}{3}=1\cdot {\left(\frac{x}{3}\right)}^{-1}=\underline{\underline{\frac{3}{x}}}$    

$x\ne 0$    

 

$b)$  

$2\frac{x}{5}:4=2\frac{x}{5}\cdot \frac{1}{4}=\frac{2x}{20}=\underline{\underline{\frac{x}{10}}}$   

$x\in \mathbb{R}$  

 

$c)$  

$\frac{3}{x-1}:\frac{2}{x}=\frac{3}{x-1}\cdot \frac{x}{2}=\underline{\underline{\frac{3x}{2x-2}}}$   

$x-1\ne 0\wedge x\ne 0$  

$x\in \mathbb{R}\backslash \left\{0;1\right\}$     

 

$d)$  

$\frac{3}{x+1}:\frac{1}{x+1}=\frac{3}{x+1}\cdot \frac{x+1}{1}=\underline{\underline{3}}$    

$x+1\ne 0\Rightarrow x\ne -1$  

 

$e)$  

$\frac{1}{x+6}:\frac{3}{2x+12}=\frac{1}{x+6}\cdot \frac{2x+12}{3}=\frac{1}{x+6}\cdot \frac{2\left(x+6\right)}{3}=\underline{\underline{\frac{2}{3}}}$   

$x+6\ne 0\wedge 2x+12\ne 0$  

$x\ne -6\wedge x\ne -6$  

 

$f)$  

$\frac{4}{x+2}:\frac{1}{4-x^2}=\frac{4}{x+2}\cdot \frac{4-x^2}{1}=\frac{4}{x+2}\cdot \left(2-x\right)\left(2+x\right)=\underline{\underline{8-4x}}$   

$x+2\ne 0\wedge 4-x^2\ne 0$  

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

$x\in \mathbb{R}\backslash \left\{-2;2\right\}$  

 

$g)$  

$\frac{5x}{x^2+10x+25}:\frac{10}{x^2-25}=\frac{5x}{x^2+10x+25}\cdot \frac{x^2-25}{10}=$  

$=\frac{5x}{{\left(x+5\right)}^2}\cdot \frac{\left(x-5\right)\left(x+5\right)}{10}=\frac{5x\left(x-5\right)}{10\left(x+5\right)}=\underline{\underline{\frac{x^2-5x}{2x+10}}}$   

$x^2+10x+25\ne 0\wedge x^2-25\ne 0$         

${\left(x+5\right)}^2\ne 0\wedge \left(x-5\right)\left(x+5\right)\ne 0$  

$x\in \mathbb{R}\backslash \left\{-5;5\right\}$  

 

$h)$  

$\frac{3x}{9x^2-81}:\frac{1}{9x-27}=\frac{3x}{9x^2-81}\cdot \left(9x-27\right)=$  

$=\frac{3x}{9\left(x-3\right)\left(x+3\right)}\cdot 9\left(x-3\right)=\underline{\underline{\frac{3x}{x+3}}}$    

$9x^2-81\ne 0\wedge 9x-27\ne 0$      

$9\left(x-3\right)\left(x+3\right)\ne 0\wedge 9\left(x-3\right)\ne 0$  

$x\in \mathbb{R}\backslash \left\{-3;3\right\}$