$a)$  

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

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

 

$D:$  

$x+1\ne 0\Rightarrow x\ne -1$   

$x-1\ne 0\Rightarrow x\ne 1$  

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

 

$b)$  

$\left(\frac{2x-3}{x+1}-\frac{2x+3}{x}\right)\cdot \frac{x^2+x}{4x^2-9}=$  

$=\frac{\left(2x-3\right)x-\left(2x+3\right)\left(x+1\right)}{x\left(x+1\right)}\cdot \frac{x\left(x+1\right)}{4x^2-9}=$   

$=\frac{2x^2-3x-2x^2-2x-3x-3}{4x^2-9}=\frac{-8x-3}{4x^2-9}$  

 

$D:$  

$x+1\ne 0\Rightarrow x\ne -1$  

$x\ne 0$  

$4x^2-9\ne 0\Rightarrow x\ne \frac{3}{2}\wedge x\ne -\frac{3}{2}$  

$D=\mathbb{R}\backslash \left\{-\frac{3}{2};-1;0;\frac{3}{2}\right\}$        

 

 

$c)$  

$\left(\frac{x^2+4x}{x^2-9}-\frac{x+4}{x-3}\right)\cdot \frac{x^2-6x+9}{x+4}=$    

$=\frac{\left(x^2+4x\right)\left(x-3\right)-\left(x+4\right)\left(x^2-9\right)}{\left(x^2-9\right)\left(x-3\right)}\cdot \frac{{\left(x-3\right)}^2}{x+4}=$  

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

 

$D:$  

$x^2-9\ne 0\Rightarrow x\ne 3\wedge x\ne -3$  

$x-3\ne 0\Rightarrow x\ne 3$  

$x+4\Rightarrow x\ne -4$  

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

 

$d)$  

$\left(\frac{x-2}{x+4}-\frac{x+2}{x-4}\right)\cdot \frac{x^2-16}{x^2-4}=$  

$=\frac{\left(x-2\right)\left(x-4\right)-\left(x+2\right)\left(x+4\right)}{x^2-16}\cdot \frac{x^2-16}{x^2-4}=$  

$=\frac{x^2-6x+8-x^2-6x-8}{x^2-4}=\frac{-12x}{x^2-4}$  

 

$D:$  

$x+4\ne 0\Rightarrow x\ne -4$  

$x-4\ne 0\Rightarrow x\ne 4$  

$x^2-4\ne \Rightarrow x\ne -2\wedge x\ne 2$  

$D=\mathbb{R}\backslash \left\{-4;-2;2;4\right\}$