Zad. 5.23
a)
[tex]a = \frac{1}{\sqrt{3} - 1} =\frac{1}{\sqrt{3} - 1} \cdot \frac{\sqrt{3} +1}{\sqrt{3} + 1}= \frac{\sqrt{3} +1}{3-1}= \frac{\sqrt{3} +1}{2} \\\\ b = \frac{2}{\sqrt{5} -1} -\frac{2}{\sqrt{5} +1} =\frac{2(\sqrt{5} +1)}{(\sqrt{5} -1)(\sqrt{5} +1)} -\frac{2(\sqrt{5} -1)}{(\sqrt{5} -1)(\sqrt{5} +1)}=\frac{2(\sqrt{5} +1)-2(\sqrt{5} -1)}{(\sqrt{5} -1)(\sqrt{5} +1)} = \\ = \frac{2\sqrt{5} +2-2\sqrt{5}+2}{5-1} =\frac{4}{4} = 1[/tex]
[tex]\frac{\sqrt{3} +1}{2} > 1[/tex]
Zatem: a > b
b)
[tex]a = \frac{(\sqrt{2}-3)^2}{2} = \frac{2-6\sqrt{2} +9}{2} = \frac{11-6\sqrt{2}}{2} \\\\ b = \frac{1}{2-\sqrt{2}} = \frac{1}{2-\sqrt{2}} \cdot \frac{2+\sqrt{2}}{2+\sqrt{2}} =\frac{2+\sqrt{2}}{4-2} =\frac{2+\sqrt{2}}{2}[/tex]
[tex]\frac{11-6\sqrt{2}}{2} <\frac{2+\sqrt{2}}{2}[/tex]
Zatem: a < b
