a)

Korzystamy ze wzoru skróconego mnożenia na różnicę kwadratów $a^2-b^2=\left(a-b\right)\left(a+b\right)$:

$\frac{6}{3-2\sqrt{3}}=\frac{6}{3-2\sqrt{3}}\cdot \frac{3+2\sqrt{3}}{3+2\sqrt{3}}=\frac{6\left(3+2\sqrt{3}\right)}{\left(3-2\sqrt{3}\right)\left(3+2\sqrt{3}\right)}=\frac{18+12\sqrt{3}}{3^2-{\left(2\sqrt{3}\right)}^2}=$

$=\frac{18+12\sqrt{3}}{9-4\cdot 3}=\frac{18+12\sqrt{3}}{9-12}=\frac{18+12\sqrt{3}}{-3}=-\frac{{\cancel{6}}^2\left(3+2\sqrt{3}\right)}{{\cancel{3}}_1}=\boxed{-2\left(3+2\sqrt{3}\right)}$

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b)

Korzystamy ze wzoru skróconego mnożenia na różnicę kwadratów $a^2-b^2=\left(a-b\right)\left(a+b\right)$:

$\frac{\sqrt{6}}{2\sqrt{3}+\sqrt{2}}=\frac{\sqrt{6}}{2\sqrt{3}+\sqrt{2}}\cdot \frac{2\sqrt{3}-\sqrt{2}}{2\sqrt{3}-\sqrt{2}}=\frac{\sqrt{6}\left(2\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{3}+\sqrt{2}\right)\left(2\sqrt{3}-\sqrt{2}\right)}=\frac{2\sqrt{18}-\sqrt{12}}{{\left(2\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}=$

$=\frac{2\sqrt{9\cdot 2}-\sqrt{4\cdot 3}}{4\cdot 3-2}=\frac{2\cdot 3\sqrt{2}-2\sqrt{3}}{12-2}=\frac{6\sqrt{2}-2\sqrt{3}}{10}=\frac{{\cancel{2}}^1\left(3\sqrt{2}-\sqrt{3}\right)}{{\cancel{10}}_5}=\boxed{\frac{3\sqrt{2}-\sqrt{3}}{5}}$

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c)

Korzystamy ze wzoru skróconego mnożenia na różnicę kwadratów $a^2-b^2=\left(a-b\right)\left(a+b\right)$:

$\frac{2}{\sqrt{2}+\sqrt{3}-1}=\frac{2}{\sqrt{2}+\sqrt{3}-1}\cdot \frac{\sqrt{2}+\sqrt{3}+1}{\sqrt{2}+\sqrt{3}+1}=$

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

Korzystamy ze wzoru skróconego mnożenia na kwadrat sumy ${\left(a+b\right)}^2=a^2+2ab+b^2$:

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

$=\frac{2\sqrt{2}+2\sqrt{3}+2}{2\sqrt{6}+4}=\frac{{\cancel{2}}^1\left(\sqrt{2}+\sqrt{3}+1\right)}{{\cancel{2}}_1\left(\sqrt{6}+2\right)}=\frac{\sqrt{2}+\sqrt{3}+1}{\sqrt{6}+2}=$

Jeszcze raz korzystamy ze wzoru skróconego mnożenia na różnicę kwadratów $a^2-b^2=\left(a-b\right)\left(a+b\right)$:

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

$=\frac{\sqrt{12}-2\sqrt{2}+\sqrt{18}-2\sqrt{3}+\sqrt{6}-2}{{\left(\sqrt{6}\right)}^2-2^2}=\frac{\sqrt{4\cdot 3}-2\sqrt{2}+\sqrt{9\cdot 2}-2\sqrt{3}+\sqrt{6}-2}{6-4}=$

$=\frac{2\sqrt{3}-2\sqrt{2}+3\sqrt{2}-2\sqrt{3}+\sqrt{6}-2}{2}=\frac{\sqrt{2}+\sqrt{6}-2}{2}=\boxed{\frac{\sqrt{6}+\sqrt{2}-2}{2}}$

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d)

Korzystamy ze wzoru skróconego mnożenia na różnicę kwadratów $a^2-b^2=\left(a-b\right)\left(a+b\right)$:

$\frac{1}{\sqrt{3}+\sqrt{2}+2}=\frac{1}{\sqrt{3}+\sqrt{2}+2}\cdot \frac{\sqrt{3}+\sqrt{2}-2}{\sqrt{3}+\sqrt{2}-2}=$

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

Korzystamy ze wzoru skróconego mnożenia na kwadrat sumy ${\left(a+b\right)}^2=a^2+2ab+b^2$:

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

Jeszcze raz korzystamy ze wzoru skróconego mnożenia na różnicę kwadratów $a^2-b^2=\left(a-b\right)\left(a+b\right)$:

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

$=\frac{2\sqrt{18}-\sqrt{3}+2\sqrt{12}-\sqrt{2}-4\sqrt{6}+2}{{\left(2\sqrt{6}\right)}^2-1^2}=\frac{2\sqrt{9\cdot 2}-\sqrt{3}+2\sqrt{4\cdot 3}-\sqrt{2}-4\sqrt{6}+2}{4\cdot 6-1}=$

$=\frac{2\cdot 3\sqrt{2}-\sqrt{3}+2\cdot 2\sqrt{3}-\sqrt{2}-4\sqrt{6}+2}{24-1}=\frac{6\sqrt{2}-\sqrt{3}+4\sqrt{3}-\sqrt{2}-4\sqrt{6}+2}{23}=$

$=\frac{5\sqrt{2}+3\sqrt{3}-4\sqrt{6}+2}{23}=\boxed{\frac{2+5\sqrt{2}+3\sqrt{3}-4\sqrt{6}}{23}}$