a)

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

$=25-7=18$ 

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b)

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

$=36-3=33$ 

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c)

$\left(\sqrt{5}+1\right)\left(1-\sqrt{5}\right)=\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)=$ 

$=1^2-{\left(\sqrt{5}\right)}^2=1-5=-4$ 

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d)

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

$=\frac{1}{9}-\frac{6}{4}=\frac{1}{9}-\frac{3}{2}=\frac{2}{18}-\frac{27}{18}=$ 

$=-\frac{25}{18}=-1\frac{7}{18}$

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e)

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

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

$=12-\frac{1}{4}=11\frac{3}{4}$ 

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f)

$\left(2\sqrt{2}-3\right)\left(3+2\sqrt{2}\right)=$ 

$=\left(2\sqrt{2}-3\right)\left(2\sqrt{2}+3\right)={\left(2\sqrt{2}\right)}^2-3^2=$ 

$=4\cdot 2-9=8-9=-1$

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g)

$\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)={\left(\sqrt{2}\right)}^2-{\left(\sqrt{3}\right)}^2=$ 

$=2-3=-1$ 

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h)

$\left(\sqrt{2}+\sqrt{6}\right)\left(\sqrt{6}-\sqrt{2}\right)=$ 

$=\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{6}-\sqrt{2}\right)=$

$={\left(\sqrt{6}\right)}^2-{\left(\sqrt{2}\right)}^2=6-2=4$ 

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i)

$\left(\sqrt{5}+\frac{\sqrt{2}}{2}\right)\left(\sqrt{5}-\frac{\sqrt{2}}{2}\right)=$ 

$={\left(\sqrt{5}\right)}^2-{\left(\frac{\sqrt{2}}{2}\right)}^2=5-\frac{2}{4}=5-\frac{1}{2}=4\frac{1}{2}$