a)
[tex](\sqrt{3}+1)(2\sqrt{3}-2) = \sqrt{3}\cdot2\sqrt{3}+\sqrt{3}\cdot(-2) + 1\cdot2\sqrt{3}+1\cdot(-2) =\\\\=2\sqrt{3\cdot3} -2\sqrt{3}+2\sqrt{3}-2 = 2\cdot3-2 = 6-2 = 4[/tex]
b)
[tex](\sqrt{2}+\sqrt{3})(\sqrt{5}+1) = \sqrt{2}\cdot\sqrt{5}+\sqrt{2}\cdot1 + \sqrt{3}\cdot\sqrt{5}+\sqrt{3}\cdot1=\\\\=\sqrt{2\cdot5}+\sqrt{2}+\sqrt{3\cdot5}+\sqrt{3} = \sqrt{10}+\sqrt{2}+\sqrt{15}+\sqrt{3}[/tex]
c)
[tex](\sqrt{2}+1)(5-2\sqrt{2}) = \sqrt{2}\cdot5+\sqrt{2}\cdot(-2\sqrt{2}+1\cdot5+1\cdot(-2\sqrt{2}) =\\\\=5\sqrt{2}-2\cdot2+5-2\sqrt{2} = 5\sqrt{2}-2\sqrt{2}-4+5=3\sqrt{2}+1[/tex]
d)
[tex](2+\sqrt{2})(2\sqrt{2}+3) = 2\cdot2\sqrt{2}+2\cdot3 +\sqrt{2}\cdot2\sqrt{2}+\sqrt{2}\cdot3=4\sqrt{2}+6+2\cdot2+3\sqrt{2}=\\\\=7\sqrt{2}+6+4 = 7\sqrt{2}+10[/tex]
