Răspuns:
6) a. [tex]\sqrt{5 + \sqrt{24} } =\frac{\sqrt{5+\sqrt{25-24} } }{\sqrt{2} } +\frac{\sqrt{5-\sqrt{25-24} } }{\sqrt{2} }=\sqrt{3} +\sqrt{2}[/tex]
Analog [tex]\sqrt{5+\sqrt{24} } =\sqrt{3} -\sqrt{2}[/tex]
notam cu E ecuatia initiala
=> E=[tex](\sqrt{3}+\sqrt{2}-\sqrt{3} +\sqrt{2} )^{2} =(2\sqrt{2} )^{2} =8[/tex] => fals
b) notam cu E ecuatia initiala
E=[tex]\frac{(2,3-1,3)(2,3^{2}+2,3*1,3+1,3^{2} )}{2,3^{2}+2,3*1,3+1,3^{2} } =1[/tex] => adevarat
7) rationalizam
E=[tex]\frac{\sqrt{3} }{3} +\frac{1-\sqrt{3} }{-2} +\frac{6-2\sqrt{3} }{6} =\frac{2\sqrt{3}-3+3\sqrt{3} +6-2\sqrt{3} }{6} =\frac{1+\sqrt{3} }{2}[/tex]
