$a)x-2\ne 0\wedge x-3\ne 0$  

$x\ne 2\wedge x\ne 3$  

$D=\mathbb{R}-\left\{2;3\right\}$  

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

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

$b)6x-3\ne 0\wedge x-4\ne 0$   

$6x\ne 3\wedge x\ne 4$  

$x\ne \frac{3}{6}\wedge x\ne 4$  

$x\ne \frac{1}{2}\wedge x\ne 4$  

$D=\mathbb{R}-\left\{\frac{1}{2};4\right\}$  

$\frac{2x^2-8x}{6x-3}\cdot \frac{2x-1}{x-4}=$   $\frac{2x\left(x-4\right)}{6{\sout{\left(x-\frac{1}{2}\right)}}^1}\cdot \frac{2{\sout{\left(x-\frac{1}{2}\right)}}^1}{x-4}=$  

$=\frac{2x{\sout{\left(x-4\right)}}^1}{6}\cdot \frac{2}{{\sout{x-4}}^1}=$   $\frac{2x\cdot 2}{6}=\frac{2x}{3}$  

$c)x^3-4x^2\ne 0\wedge 2x+6\ne 0$   

$x^2\left(x-4\right)\ne 0\wedge 2x\ne -6$  

$x^2\ne 0\wedge x-4\ne 0\wedge x\ne -3$  

$x\ne 0\wedge x\ne 4\wedge x\ne -3$  

$D=\mathbb{R}-\left\{-3;0;4\right\}$  

Licznik pierwszego ułamka to wzór skróconego mnożenia.

Rozłóżmy licznik drugiego mianownika na czynniki:

$x^2-3x-4$  

$\Delta ={\left(-3\right)}^2-4\cdot 1\cdot \left(-4\right)=9+16=25$  

$\sqrt{\Delta }=\sqrt{25}=5$   

$x_1=\frac{3-5}{2}=-\frac{2}{2}=-1$          

$x_2=\frac{3+5}{2}=\frac{8}{2}=4$  

$x^2-3x-4=\left(x+1\right)\left(x-4\right)$  

Wracamy do mnożenia:

$\frac{x^2+6x+9}{x^3-4x^2}\cdot \frac{x^2-3x-4}{2x+6}=$   $\frac{{\left(x+3\right)}^2}{x^2{\sout{\left(x-4\right)}}^1}\cdot \frac{\left(x+1\right){\sout{\left(x-4\right)}}^1}{2\left(x+3\right)}=$  

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

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

$=\frac{x^2+x+3x+3}{2x^2}=$   $\frac{x^2+4x+3}{2x^2}$