$a)16x^7\ne 0\iff x^7\ne 0\iff x\ne 0$  
$D=\mathbb{R}-\left\{0\right\}$    

$\frac{2x^2}{16x^7}=\frac{2x^2\cdot 1}{2x^2\cdot 8x^5}=$   $\frac{1}{8x^5}$       

$b){\left(1-x\right)}^4\ne 0\iff \left(1-x\right)\ne 0\iff x\ne 1$  

$D=\mathbb{R}-\left\{1\right\}$  

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

$c)x^2-5^2\ne 0\iff \left(x-5\right)\left(x+5\right)\ne 0\iff \left(x-5\ne 0\wedge x+5\ne 0\right)\iff \left(x\ne 5\wedge x\ne -5\right)$  

$D=\mathbb{R}-\left\{-5;5\right\}$  

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

$d)x^2-18x+81\ne 0\iff {\left(x-9\right)}^2\ne 0\iff x-9\ne 0\iff x\ne 9$  

$D=\mathbb{R}-\left\{9\right\}$  

$\frac{x^2-81}{x^2-18x+81}=\frac{\left(x-9\right)\cdot \left(x+9\right)}{{\left(x-9\right)}^2}=$   $\frac{x+9}{x-9}$  

$e)2x^2-7x+3\ne 0$  

$\Delta ={\left(-7\right)}^2-4\cdot 2\cdot 3=$   $49-24=25$  

$\sqrt{\Delta }=5$  

$x\ne \frac{7-5}{2\cdot 2}\wedge x\ne \frac{7+5}{2\cdot 2}$  

$x\ne \frac{1}{2}\wedge x\ne 3$  

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

Przy okazji wypisywania założeń mamy miejsca zerowe mianownika, możemy więc rozłożyć mianownik na czynniki: 

$2x^2-7x+3=2\cdot \left(x-\frac{1}{2}\right)\cdot \left(x-3\right)=\left(2x-1\right)\cdot \left(x-3\right)$   

Teraz rozłożymy licznik na czynniki: 

$x^2-5x+6$  

$\Delta ={\left(-5\right)}^2-4\cdot 1\cdot 6=25-24=1$  

$\sqrt{\Delta }=1$  

$x=\frac{5-1}{2}=2\vee x=\frac{5+1}{2}=3$  

$x^2-5x+6=\left(x-2\right)\cdot \left(x-3\right)$  

Teraz wracamy do skracania ułamka: 

$\frac{x^2-5x+6}{2x^2-7x+3}=\frac{\left(x-2\right)\cdot \left(x-3\right)}{\left(2x-1\right)\cdot \left(x-3\right)}=$   $\frac{x-2}{2x-1}$  

$f)x^3+x^2+5x+5\ne 0$  

$x^2\left(x+1\right)+5\left(x+1\right)\ne 0$  

$\left(x+1\right)\left(x^2+5\right)\ne 0$  

$x+1\ne 0\wedge x^2+5\ne 0$  

$x\ne -1\wedge x\in \mathbb{R}$   (kwadrat liczby jest nieujemny, jeśli powiększymy go jescze o 5 to na pewno nie otrzymamy 0)

$D=\mathbb{R}-\left\{-1\right\}$  

$\frac{x\left(x^2+2x+1\right)}{x^3+x^2+5x+5}=$   $\frac{x\cdot {\left(x+1\right)}^2}{\left(x+1\right)\cdot \left(x^2+5\right)}=$   $\frac{x\cdot \left(x+1\right)}{x^2+5}$