$a)$

$\{\begin{array}{l}x+3\ne 0\\ x^2\ne 0\end{array}\Rightarrow \{\begin{array}{l}x\ne -3\\ x\ne 0\end{array}\Rightarrow D=\mathbb{R}\backslash \left\{-3,0\right\}$

 

$\frac{6x}{\sout{x+3}}\cdot \frac{\sout{x+3}}{x^2}=\frac{6x}{x^2}=\frac{6}{x}$

 

$b)$

$\{\begin{array}{l}4x\ne 0\\ 2x+4\ne 0\end{array}\Rightarrow \{\begin{array}{l}x\ne 0\\ x\ne -2\end{array}\Rightarrow D=\mathbb{R}\backslash \left\{-2,0\right\}$

 

$\frac{\sout{2x+4}}{4x}\cdot \frac{8x^2}{\sout{2x+4}}=\frac{8x^2}{4x}=2x$

 

 

$c)$

$\{\begin{array}{l}x-3\ne 0\\ x\ne 0\end{array}\Rightarrow \{\begin{array}{l}x\ne 3\\ x\ne 0\end{array}\Rightarrow D=\mathbb{R}\backslash \left\{0,3\right\}$

 

$\frac{x+1}{x-3}\cdot \frac{2x-6}{x}=\frac{x+1}{\sout{x-3}}\cdot \frac{2\cdot \sout{\left(x-3\right)}}{x}=\frac{2\left(x+1\right)}{x}=\frac{2x+2}{x}$

 

 

$d)$

$\{\begin{array}{l}18x\ne 0\\ 4-x\ne 0\end{array}\Rightarrow \{\begin{array}{l}x\ne 0\\ x\ne 4\end{array}\Rightarrow D=\mathbb{R}\backslash \left\{0,4\right\}$

 

$\frac{x-4}{18x}\cdot \frac{12}{4-x}=\frac{\sout{x-4}}{18x}\cdot \frac{-12}{\sout{x-4}}=\frac{-12}{18x}=-\frac{2}{3x}$

 

 

$e)$

$\{\begin{array}{l}x+1\ne 0\\ x-5\ne 0\end{array}\Rightarrow \{\begin{array}{l}x\ne -1\\ x\ne 5\end{array}\Rightarrow D=\mathbb{R}\backslash \left\{-1,5\right\}$

 

$\frac{2x-10}{x+1}\cdot \frac{3x+3}{x-5}=\frac{2\sout{\left(x-5\right)}}{x+1}\cdot \frac{3\left(x+1\right)}{\sout{x-5}}=\frac{2}{\sout{x+1}}\cdot 3\cdot \sout{\left(x+1\right)}=6$

 

 

$f)$

$\{\begin{array}{l}x-2\ne 0\\ 3x+3\ne 0\end{array}\Rightarrow \{\begin{array}{l}x\ne 2\\ x\ne -1\end{array}\Rightarrow D=\mathbb{R}\backslash \left\{-1,2\right\}$

 

$\frac{2x+2}{x-2}\cdot \frac{x-3}{3x+3}=\frac{2\cdot \sout{\left(x+1\right)}}{x-2}\cdot \frac{x-3}{3\cdot \sout{\left(x+1\right)}}=\frac{2\left(x-3\right)}{3\left(x-2\right)}=\frac{2x-6}{3x-6}$