a)

Wykonujemy działanie i otrzymujemy 

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

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

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

Założenie:

$\left(x-2\right)\left(x+2\right)\ne 0$  

$x-2\ne 0\wedge x+2\ne 0$  

$x\ne 2x\ne -2$  

czyli 

$x\in \mathbb{R}\backslash \left\{-2,2\right\}$  

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b)

Wykonujemy działanie i otrzymujemy 

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

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

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

Założenie:

$\left(x-3\right)\left(x+3\right)\ne 0$  

$x-3\ne 0\wedge x+3\ne 0$  

$x\ne 3x\ne -3$  

czyli 

$x\in \mathbb{R}\backslash \left\{-3,3\right\}$  

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c)

Wykonujemy działanie i otrzymujemy 

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

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

Założenie:

${\left(2x+1\right)}^2\ne 0$  

$2x+1\ne 0$

$2x\ne -1$   

$x\ne -\frac{1}{2}$  

czyli 

$x\in \mathbb{R}\backslash \left\{-\frac{1}{2}\right\}$  

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d)

Wykonujemy działanie i otrzymujemy 

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

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

$=\frac{x^2-1-4x-6}{{\left(2x+3\right)}^2}=\frac{x^2-4x-7}{{\left(2x+3\right)}^2}$  

Założenie:

${\left(2x+3\right)}^2\ne 0$  

$2x+3\ne 0$  

$2x\ne -3$  

$x\ne -\frac{3}{2}$  

czyli 

$x\in \mathbb{R}\backslash \left\{-\frac{3}{2}\right\}$  

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e)

Wykonujemy działanie i otrzymujemy 

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

$=\frac{x^2-16}{\left(x-1\right)\left(x-4\right)}-\frac{x^2-1}{\left(x-1\right)\left(x-4\right)}=\frac{x^2-16-\left(x^2-1\right)}{\left(x-1\right)\left(x-4\right)}=$  

$=\frac{x^2-16-x^2+1}{\left(x-1\right)\left(x-4\right)}=\frac{-15}{\left(x-1\right)\left(x-4\right)}$  

Założenie:

$\left(x-1\right)\left(x-4\right)\ne 0$  

$x-1\ne 0\wedge x-4\ne 0$  

$x\ne 1x\ne 4$  

czyli 

$x\in \mathbb{R}\backslash \left\{1,4\right\}$  

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f)

Wykonujemy działanie i otrzymujemy 

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

$=\frac{2x+x^2+10+5x}{\left(x-5\right)\left(2+x\right)}-\frac{2x-10-x^2+5x}{\left(x-5\right)\left(2+x\right)}=$  

$=\frac{x^2+7x+10-\left(-x^2+7x-10\right)}{\left(x-5\right)\left(2+x\right)}=$  

$=\frac{x^2+7x+10+x^2-7x+10}{\left(x-5\right)\left(2+x\right)}=\frac{2x^2+20}{\left(x-5\right)\left(2+x\right)}$  

Założenie:

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

$x-5\ne 0\wedge 2+x\ne 0$  

$x\ne 5x\ne -2$  

czyli 

$x\in \mathbb{R}\backslash \left\{-2,5\right\}$