$\text{a)}\frac{4}{x-2}-\frac{5}{x+2}=\frac{20}{x^2-4}$  

Określamy dziedzinę D równania.  

$x-2\ne 0{\text{i}}^{}x+2\ne 0\text{i}{x}^2-4\ne 0$  

$x\ne 2\text{i}x\ne -2\text{i}\left(x-2\right)\left(x+2\right)\ne 0$      

$x\ne 2\text{i}x\ne -2\text{i}x\ne 2\text{i}x\ne -2$  

Zatem: 

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

 

Rozwiązujemy równanie. 

$\frac{4\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{20}{\left(x+2\right)\left(x-2\right)}$  

$\frac{4\left(x+2\right)-5\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{20}{\left(x+2\right)\left(x-2\right)}$  

$4\left(x+2\right)-5\left(x-2\right)=20$  

$4x+8-5x+10=20$  

$-x+18=20\mid -18$  

$-x=2\mid \cdot \left(-1\right)$  

$x=-2\notin D$  

Zatem równanie nie ma rozwiązań. 

 

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$\text{b)}\frac{5x^2-32x+3}{x^2-4x+3}=2-\frac{9-3x}{x-1}$  

Określamy dziedzinę D równania. 

$x^2-4x+3\ne 0\text{i}x-1\ne 0$      

$x^2-x-3x+3\ne 0\text{i}x\ne 1$     

$x\left(x-1\right)-3\left(x-1\right)\ne 0\text{i}x\ne 1$  

$\left(x-3\right)\left(x-1\right)\ne 0\text{i}x\ne 1$    

$x-3\ne 0\text{i}x-1\ne 0\text{i}x\ne 1$   

$x\ne 3\text{i}x\ne 1\text{i}x\ne 1$  

Zatem: 

$D=\mathbb{R}\backslash \left\{1,3\right\}$      

 

Rozwiązujemy równanie. 

$\frac{5x^2-32x+3}{x^2-4x+3}=2-\frac{9-3x}{x-1}$  

$\frac{5x^2-32x+3}{\left(x-3\right)\left(x-1\right)}=\frac{2\left(x-3\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}-\frac{-3\left(x-3\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}$  

$\frac{5x^2-32x+3}{\left(x-3\right)\left(x-1\right)}=\frac{2\left(x-3\right)\left(x-1\right)+3{\left(x-3\right)}^2}{\left(x-1\right)\left(x-3\right)}$  

$5x^2-32x+3=2\left(x^2-4x+3\right)+3\left(x^2-6x+9\right)$    

$5x^2-32x+3=2x^2-8x+6+3x^2-18x+27$  

$5x^2-32x+3=5x^2-26x+33\mid -5x^2$  

$-32x+3=-26x+33\mid +26x$  

$-6x+3=33\mid -3$  

$-6x=30\mid :\left(-6\right)$  

$x=-5\in D$  

Zatem: 

${\mathbf{x}}_0=-5$