Zad. 7
a)
[tex](\sqrt{4-\sqrt{7}} +\sqrt{4+\sqrt{7}})^2 = (\sqrt{4-\sqrt{7}})^2+2 \cdot \sqrt{4-\sqrt{7}} \cdot \sqrt{4+\sqrt{7}}+(\sqrt{4+\sqrt{7}})^2 = \\ =4-\sqrt{7}+2 \cdot \sqrt{(4-\sqrt{7}) \cdot (4+\sqrt{7})}+4+\sqrt{7}} =8 + 2\cdot \sqrt{4^2-(\sqrt{7})^2}=\\ =8 + 2\cdot \sqrt{16-7}=8 + 2\cdot \sqrt{9}= 8 + 2\cdot 3 = 8 +6 = 14[/tex]
b)
[tex](\sqrt{2+\sqrt{5}} -\sqrt{\sqrt{5}-2})^2 =(\sqrt{2+\sqrt{5}})^2-2 \cdot \sqrt{2+\sqrt{5}} \cdot \sqrt{\sqrt{5}-2}+(\sqrt{\sqrt{5}-2})^2 = \\ =2+\sqrt{5}-2 \cdot \sqrt{(2+\sqrt{5}) \cdot (\sqrt{\sqrt{5}-2})}+\sqrt{5}-2 =2\sqrt{5} -2 \cdot \sqrt{(\sqrt{5})^2-2^2}= \\ = 2\sqrt{5} -2 \cdot \sqrt{5-4} =2\sqrt{5} -2 \cdot \sqrt{1} =2\sqrt{5} -2 \cdot 1=2\sqrt{5} -2= 2 \cdot (\sqrt{5} -1)[/tex]
c)
[tex](\sqrt{5-2\sqrt{6}} +\sqrt{5+2\sqrt{6}})^2 =(\sqrt{5-2\sqrt{6}})^2 +2 \cdot \sqrt{5-2\sqrt{6}} \cdot \sqrt{5+2\sqrt{6}} +(\sqrt{5+2\sqrt{6}})^2 =\\ =5-2\sqrt{6}+2 \cdot \sqrt{(5-2\sqrt{6}) \cdot (5+2\sqrt{6})}+5+2\sqrt{6} =10+2 \cdot \sqrt{5^2-(2\sqrt{6})^2}=\\ =10+2 \cdot \sqrt{25-4 \cdot 6}=10+2 \cdot \sqrt{25-24}=10+2 \cdot \sqrt{1}=10+2 \cdot 1=10+2=12[/tex]
d)
[tex](\sqrt{\sqrt{7} +\sqrt{3}} -\sqrt{\sqrt{7} -\sqrt{3}})^2 =(\sqrt{\sqrt{7} +\sqrt{3}})^2 -2 \cdot \sqrt{\sqrt{7} +\sqrt{3}} \cdot \sqrt{\sqrt{7} -\sqrt{3}}+ \\ +(\sqrt{\sqrt{7} -\sqrt{3}})^2 =\sqrt{7} +\sqrt{3} -2 \cdot \sqrt{(\sqrt{7} +\sqrt{3}) \cdot (\sqrt{7} -\sqrt{3})}+\sqrt{7} -\sqrt{3} = \\ = 2\sqrt{7} -2 \cdot \sqrt{(\sqrt{7})^2-(\sqrt{3})^2}= 2\sqrt{7} -2 \cdot \sqrt{7-3}=2\sqrt{7} -2 \cdot \sqrt{4}=2\sqrt{7} -2 \cdot 2= \\ = 2\sqrt{7} -4 = 2 \cdot (\sqrt{7} -2)[/tex]
