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

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

Wyznaczamy dziedzinę D wyrażenia. 

$x\ne 0$  

Zatem: 

$D=\mathbb{R}\backslash \left\{0\right\}$  

 

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

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

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

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

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

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

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

$=\frac{2x^2-5x-1}{x{\left(x-1\right)}^2}$  

Wyznaczamy dziedzinę D wyrażenia. 

$x\ne 0\text{i}x-1\ne 0$  

$x\ne 0\text{i}x\ne 1$  

Zatem: 

$D=\mathbb{R}\backslash \left\{0,1\right\}$  

 

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

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

$=\frac{2x^2+x-18}{x^2-4}$  

Wyznaczamy dziedzinę D wyrażenia. 

$x-2\ne 0\text{i}x+2\ne 0$  

$x\ne 2\text{i}x\ne -2$  

Zatem: 

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

 

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

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

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

Wyznaczamy dziedzinę D wyrażenia. 

$x+2\ne 0$  

$x\ne -2$  

Zatem: 

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

 

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

$=\frac{2x^2+6x+x+3}{x^2-9}-\frac{x^2-3x-2x+6}{x^2-9}-\frac{x+1}{x^2-9}=$  

$=\frac{2x^2+7x+3-\left(x^2-5x+6\right)-\left(x+1\right)}{x^2-9}=$  

$=\frac{2x^2+7x+3-x^2+5x-6-x-1}{x^2-9}=\frac{x^2+11x-4}{x^2-9}$  

Wyznaczamy dziedzinę D wyrażenia. 

$x-3\ne 0\text{i}x+3\ne 0$  

$x\ne 3\text{i}x\ne -3$  

Zatem: 

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

Uwaga!!! W odpowiedziach podano błędne rozwiązanie. W liczniku ułamka powinno być 4 a nie 3. 

 

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$\text{f)}\frac{y}{y-2}+\frac{8y^2-4y}{8-y^3}+\frac{4}{y^2+2y+4}=$   

$=\frac{y}{y-2}+\frac{8y^2-4y}{\left(2-y\right)\left(4+2y+y^2\right)}+\frac{4}{y^2+2y+4}=$  

$=-\frac{y\left(y^2+2y+4\right)}{\left(2-y\right)\left(y^2+2y+4\right)}+\frac{8y^2-4y}{\left(2-y\right)\left(y^2+2y+4\right)}+\frac{4\left(2-y\right)}{\left(y^2+2y+4\right)\left(2-y\right)}=$        

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

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

$=\frac{-y^3+6y^2-12y+8}{\left(y^2+2y+4\right)\left(2-y\right)}=\frac{-\left(y^3-6y^2+12y-8\right)}{\left(y^2+2y+4\right)\left(2-y\right)}=$      

$=\frac{-{\sout{{\left(y-2\right)}^3}}^{{\left(y-2\right)}^2}}{-\left(y^2+2y+4\right){\sout{\left(y-2\right)}}_1}=\frac{{\left(y-2\right)}^2}{y^2+2y+4}$     

Wyznaczamy dziedzinę D wyrażenia. 

$8-y^3\ne 0$  

$8\ne y^3$  

$y\ne 2$  

Zatem: 

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