W celu usunięcia niewymierności z mianownika skorzystamy ze wzoru skróconego mnożenia na różnicę kwadratów, czyli:

$a^2-b^2=\left(a-b\right)\left(a+b\right)$

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$\text{a)}\frac{1}{6+\sqrt{2}}=\frac{1\cdot \left(6-\sqrt{2}\right)}{\left(6+\sqrt{2}\right)\left(6-\sqrt{2}\right)}=\frac{6-\sqrt{2}}{36-2}=\frac{6-\sqrt{2}}{34}$

$\text{b)}\frac{2\sqrt{2}}{\sqrt{5}+4}=\frac{2\sqrt{2}\cdot \left(\sqrt{5}-4\right)}{\left(\sqrt{5}+4\right)\left(\sqrt{5}-4\right)}=\frac{2\sqrt{10}-8\sqrt{2}}{5-16}=\frac{2\sqrt{10}-8\sqrt{2}}{-11}=\frac{8\sqrt{2}-2\sqrt{10}}{11}$

$\text{c)}\frac{2}{4-3\sqrt{2}}=\frac{2\cdot \left(4+3\sqrt{2}\right)}{\left(4-3\sqrt{2}\right)\left(4+3\sqrt{2}\right)}=\frac{8+6\sqrt{2}}{16-9\cdot 2}=\frac{8+6\sqrt{2}}{16-18}=\frac{8+6\sqrt{2}}{-2}=-4-3\sqrt{2}$

$\text{d)}\frac{2\sqrt{3}-1}{2\sqrt{3}+2}=\frac{\left(2\sqrt{3}-1\right)\cdot \left(2\sqrt{3}-2\right)}{\left(2\sqrt{3}+2\right)\cdot \left(2\sqrt{3}-2\right)}=\frac{2\sqrt{3}\cdot \left(2\sqrt{3}-2\right)-1\cdot \left(2\sqrt{3}-2\right)}{4\cdot 3-4}=$

$=\frac{4\cdot 3-4\sqrt{3}-2\sqrt{3}+2}{12-4}=\frac{14-6\sqrt{3}}{8}=\frac{7-3\sqrt{3}}{4}$

$\text{e)}\frac{\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\frac{\sqrt{3}\cdot \left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\cdot \left(\sqrt{5}-\sqrt{3}\right)}=\frac{\sqrt{15}-3}{5-3}=\frac{\sqrt{15}-3}{2}$