$a)$

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

 

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

 

$x=\frac{1}{2}\in D\Rightarrow \frac{3x}{2}=\left(3\cdot \frac{1}{2}\right):2=\frac{3}{2}:2=\frac{3}{2}\cdot \frac{1}{2}=\frac{3}{4}$

 

 

 

$b)$

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

 

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

 

$x=\frac{1}{2}\in D\Rightarrow -\frac{4}{x}=-\frac{4}{\frac{1}{2}}=-4:\frac{1}{2}=-4\cdot 2=-8$

 

 

 

$c)$

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

 

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

 

$x=\frac{1}{2}\in D\Rightarrow -3x^2=-3\cdot {\left(\frac{1}{2}\right)}^2=-3\cdot \frac{1}{4}=-\frac{3}{4}$

 

 

 

$d)$

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

 

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

 

$x=\frac{1}{2}\notin D,\text{czyli wartosˊcˊ nie istnieje}$

 

 

 

$e)$

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

 

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

 

$x=\frac{1}{2}\in D\Rightarrow 2x+2=2\cdot \frac{1}{2}+2=1+2=3$

 

 

$f)$

$\{\begin{array}{l}x-4\ne 0\\ 4x+8\ne 0\end{array}\Rightarrow \{\begin{array}{l}x\ne 4\\ x\ne -2\end{array}\Rightarrow D=\mathbb{R}\backslash \left\{-2,4\right\}$

 

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

 

$x=\frac{1}{2}\in D\Rightarrow \frac{x-4}{4}=\frac{\frac{1}{2}-4}{4}=\left(-3\frac{1}{2}\right):4=-\frac{7}{2}\cdot \frac{1}{4}=-\frac{7}{8}$