a) Wyznaczamy dziedzinę

$x^2-4=0\text{lub}2x-4=0$  

$\left(x-2\right)\left(x+2\right)=0\text{lub}x-2=0$  

Stąd

$x=2\text{lub}x=-2$  

$\mathbf{D}=\mathbb{R}\backslash \left\{-2,2\right\}$

Wykonujemy działanie

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

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b) Wyznaczamy dziedzinę

$x+2=0\text{lub}{x}^2-4=0\text{lub}{x}^2-4x+4=0$  

$x=-2\text{lub}\left(x-2\right)\left(x+2\right)=0\text{lub}{\left(x-2\right)}^2=0$   

Stąd

$x=2\text{lub}x=-2$  

$\mathbf{D}=\mathbb{R}\backslash \left\{-2,2\right\}$

Wykonujemy działanie

$\frac{2}{x+2}+\frac{x-3}{x^2-4}-\frac{2x+1}{x^2-4x+4}=\frac{2}{x+2}+\frac{x-3}{\left(x-2\right)\left(x+2\right)}-\frac{2x+1}{{\left(x-2\right)}^2}=\frac{2{\left(x-2\right)}^2}{\left(x+2\right){\left(x-2\right)}^2}+\frac{\left(x-3\right)\left(x-2\right)}{\left(x+2\right){\left(x-2\right)}^2}-\frac{\left(2x+1\right)\left(x+2\right)}{\left(x+2\right){\left(x-2\right)}^2}=\frac{2\left(x^2-4x+4\right)+x^2-5x+6-\left(2x^2+5x+2\right)}{\left(x+2\right){\left(x-2\right)}^2}=\frac{2x^2-8x+8+x^2-5x+6-2x^2-5x-2}{\left(x+2\right){\left(x-2\right)}^2}=\frac{x^2-18x+12}{\left(x+2\right){\left(x-2\right)}^2}$  

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c) Wyznaczamy dziedzinę

$x^2-5x+6=0\text{lub}{x}^2+x-6=0\text{lub}{x}^2-9=0$  

$x^2-3x-2x+6=0\text{lub}{x}^2+3x-2x-6=0\text{lub}\left(x-3\right)\left(x+3\right)=0$   

$x\left(x-3\right)-2\left(x-3\right)=0\text{lub}x\left(x+3\right)-2\left(x+3\right)=0\text{lub}\left(x-3\right)\left(x+3\right)=0$   

$\left(x-3\right)\left(x-2\right)=0\text{lub}\left(x+3\right)\left(x-2\right)=0\text{lub}\left(x-3\right)\left(x+3\right)=0$   

Stąd

$x=3\text{lub}x=2\text{lub}x=-3$  

$\mathbf{D}=\mathbb{R}\backslash \left\{-3,-2,3\right\}$

Wykonujemy działanie

$\frac{2}{x^2-5x+6}-\frac{3}{x^2+x-6}+\frac{4}{x^2-9}=\frac{2}{\left(x-3\right)\left(x-2\right)}-\frac{3}{\left(x+3\right)\left(x-2\right)}+\frac{4}{\left(x-3\right)\left(x+3\right)}=\mathfrak{a}\left\{2\left(x+3\right)\right\}\left\{\left(x-3\right)\left(x+3\right)\left(x-2\right)\right\}-\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)\left(x-2\right)}+\frac{4\left(x-2\right)}{\left(x-3\right)\left(x+3\right)\left(x-2\right)}=\frac{2x+6-3x+9+4x-8}{\left(x-3\right)\left(x+3\right)\left(x-2\right)}=\frac{3x+7}{\left(x-3\right)\left(x+3\right)\left(x-2\right)}$  

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d) Wyznaczamy dziedzinę

$x^3-1=0\text{lub}\underset{\Delta <0}{x^2+x+1}=0\text{lub}x-1=0$    

Stąd

$x=1$  

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

Wykonujemy działanie

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

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e) Wyznaczamy dziedzinę

$x^3+8=0\text{lub}\underset{\Delta <0}{x^2-2x+4}=0\text{lub}x+2=0$    

Stąd

$x=-2$  

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

Wykonujemy działanie

$\frac{2}{x^3+8}-\frac{1}{x+2}-\frac{5}{x^2-2x+4}=\frac{2}{\left(x+2\right)\left(x^2-2x+4\right)}-\frac{1}{x+2}-\frac{5}{x^2-2x+4}=\frac{2}{\left(x+2\right)\left(x^2-2x+4\right)}-\frac{x^2-2x+4}{\left(x+2\right)\left(x^2-2x+4\right)}-\frac{5\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{2-x^2+2x-4-5x-10}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{-x^2-3x-12}{x^3+8}$  

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f) Wyznaczamy dziedzinę

$x^3+27=0\text{lub}\underset{\Delta <0}{x^2-3x+9}=0\text{lub}x+3=0$    

Stąd

$x=-3$  

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

Wykonujemy działanie

$\frac{1}{x^3+27}-\frac{4}{x+3}+\frac{2}{x^2-3x+9}=\frac{1}{\left(x+3\right)\left(x^2-3x+9\right)}-\frac{4}{x+3}+\frac{2}{x^2-3x+9}=\frac{1}{\left(x+3\right)\left(x^2-3x+9\right)}-\frac{4\left(x^2-3x+9\right)}{\left(x+3\right)\left(x^2-3x+9\right)}+\frac{2\left(x+3\right)}{\left(x+3\right)\left(x^2-3x+9\right)}=\frac{1-4x^2+12x-36+2x+6}{\left(x+3\right)\left(x^2-3x+9\right)}=\frac{-4x^2+14x-29}{x^3+27}$