[tex]a) \ \frac{4}{\sqrt{3}-1}=\frac{4(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{4(1+\sqrt{3})}{3-1} = \frac{4(1+\sqrt{3})}{2}=2(1+\sqrt{3}) = 2+2\sqrt{3}[/tex]
[tex]b) \ \frac{2}{2-\sqrt{3}}=\frac{2(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})} = \frac{2(2+\sqrt{3})}{4-3} = \frac{2(2+\sqrt{3})}{1} = 2(2+\sqrt{3}) = 4+2\sqrt{3}[/tex]
[tex]c) \ \frac{2}{2\sqrt{3}-4} =\frac{2}{2(\sqrt{3}-2)} = \frac{1}{\sqrt{3}-2}=\frac{\sqrt{3}+2}{(\sqrt{3}-2)(\sqrt{3}+2)} = \frac{2+\sqrt{3}}{3-4} = \frac{2+\sqrt{3}}{-1} =\\\\= -(2+\sqrt{3}) = -2-\sqrt{3}[/tex]
[tex]d) \ \frac{2\sqrt{3}}{\sqrt{3}+1}=\frac{2\sqrt{3}(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} =\frac{2(3-\sqrt{3})}{3-1} =\frac{2(3-\sqrt{3})}{2} = 3-\sqrt{3}[/tex]
Wykorzystano wzór skróconego mnożenia:
[tex](a+b)(a-b) = a^{2}-b^{2}[/tex]
