$\text{a)}$  

Zał:

$x+3\ne 0$  

$x\ne -3$  

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

 

Dla x=2 otrzymujemy:   

$\frac{2-2}{2+3}=\frac{0}{5}=0$  

Dla x=-√2 otrzymujemy:

$\frac{-\sqrt{2}-2}{-\sqrt{2}+3}=\frac{\sqrt{2}+2}{\sqrt{2}-3}=\frac{\left(\sqrt{2}+2\right)\left(\sqrt{2}+3\right)}{\left(\sqrt{2}-3\right)\left(\sqrt{2}+3\right)}=\frac{2+3\sqrt{2}+2\sqrt{2}+6}{{\sqrt{2}}^2-3^2}=\frac{8+5\sqrt{2}}{-7}=\frac{-8-5\sqrt{2}}{7}$  

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$\text{b)}$  

Zał:

$x^2-4x+2\ne 0$  

$\Delta ={\left(-4\right)}^2-4\cdot 1\cdot 2=16-8=8$  

$x_1=\frac{4-\sqrt{8}}{2}=\frac{4-2\sqrt{2}}{2}=2-\sqrt{2}$  

$x_2=\frac{4+\sqrt{8}}{2}=\frac{4+2\sqrt{2}}{2}=2+\sqrt{2}$  

$D=\mathbb{R}\text{}\left\{2-\sqrt{2},2+\sqrt{2}\right\}$  

 

Dla x=2 otrzymujemy:

$\frac{2\cdot 2-1}{2^2-4\cdot 2+2}=\frac{4-1}{4-8+2}=\frac{3}{-2}=-1\frac{1}{2}$  

Dla x=-√2 otrzymujemy:

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

$=\frac{\left(-2\sqrt{2}-1\right)\left(4-4\sqrt{2}\right)}{\left(4+4\sqrt{2}\right)\left(4-4\sqrt{2}\right)}=\frac{-8\sqrt{2}+16-4+4\sqrt{2}}{4^2-{\left(4\sqrt{2}\right)}^2}=\frac{-4\sqrt{2}+12}{16-32}=$  

$=\frac{-4\left(\sqrt{2}-3\right)}{-16}=\frac{\sqrt{2}-3}{4}$