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

$D:$  

$x^2-1\ne 0\Rightarrow x\ne 1\wedge x\ne -1$    

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

$f\left(0\right)=4$   

Wykres nr II.

 

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

$D=\mathbb{R}$   

Wykres nr III.

 

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

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

$h\left(0\right)=0$   

Wykres nr I. 

 

$a)$  

$f\left(x\right)\le 0$  

Odczytujemy z rysunku, że:

$x\in \left[-2;-1\right)\cup \left(1;2\right]$  

 

$b)$  

$g\left(x\right)\cdot h\left(x\right)\ge 0$  

$\frac{2x-x^2}{x^2-1}\cdot \frac{x^2-4}{x^2+1}\ge 0$  

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

 

$\frac{2x-x^2}{x^2-1}\cdot \frac{x^2-4}{x^2+1}\ge 0\mid \cdot \left(x^2+1\right)>0$  

$\frac{2x-x^2}{\left(x-1\right)\left(x+1\right)}\cdot \frac{x^2-4}{1}\ge 0$   

$\frac{x\left(2-x\right)\left(x-2\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)}\ge 0$    

$-x{\left(x-2\right)}^2\left(x+2\right)\left(x-1\right)\left(x+1\right)\ge 0$   

$x\in \left[-2;-1\right)\cup \left[0;1\right)\cup \left\{2\right\}$  

 

$c)$  

$\frac{h\left(x\right)}{f\left(x\right)}\ge 0$  

$\frac{\frac{2x-x^2}{x^2-1}}{\frac{x^2-4}{x^2-1}}\ge 0$   

$\frac{2x-x^2}{x^2-1}\cdot \frac{x^2-1}{x^2-4}\ge 0$  

$D:$  

$x^2-1\ne 0\Rightarrow x\ne 1\wedge x\ne -1$  

$x^2-4=\left(x-2\right)\left(x+2\right)\ne 0\Rightarrow x\ne -2\wedge x\ne 2$  

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

 

$\frac{2x-x^2}{x^2-1}\cdot \frac{x^2-1}{x^2-4}\ge 0$  

$\frac{-x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\ge 0$  

$\frac{x}{x+2}\le 0$  

$x\left(x+2\right)\le 0$  

$x_1=0$  

$x_2=-2$  

$\underline{x\in \left(-2;0\right]\backslash \left\{-1\right\}}$