Question

Problema 72, cu rezolvare daca se poate. Am uitat formula.

Answer 1

   
[tex]a= \sqrt{30-12 \sqrt{6} } +2 \sqrt{4- \sqrt{12}}- \sqrt{27-18 \sqrt{2}} = \\ \\ = \sqrt{30-12 \sqrt{6} } +2 \sqrt{4- 2\sqrt{3}}- \sqrt{27-18 \sqrt{2}} = \\ \\ = \sqrt{18+12-2\cdot 3\sqrt{2} \cdot 2\sqrt{3} } +2 \sqrt{3+1- 2\cdot\sqrt{3}\cdot 1}- \\ - \sqrt{18+9-2\cdot3\sqrt{2}\cdot3} = \\ \\ = \sqrt{18-2\cdot 3\sqrt{2} \cdot 2\sqrt{3} +12} +2 \sqrt{3- 2\cdot\sqrt{3}\cdot 1+1}- \\ - \sqrt{18-2\cdot3\sqrt{2}\cdot3+9} = [/tex]

[tex]= \sqrt{(3\sqrt{2})^2-2\cdot 3\sqrt{2} \cdot 2\sqrt{3} +(2\sqrt{3})^2} +2 \sqrt{(\sqrt{3})^2- 2\cdot\sqrt{3}\cdot 1+1^2}- \\ - \sqrt{(3\sqrt{2})^2 -2\cdot3\sqrt{2}\cdot3+3^2} = \\ \\ = \sqrt{(3\sqrt{2}-2\sqrt{3})^2} +2 \sqrt{(\sqrt{3}-1)^2} - \sqrt{(3\sqrt{2}-3)^2} = \\ \\ =3\sqrt{2}-2\sqrt{3}+2(\sqrt{3}-1 )-( 3\sqrt{2}-3)= \\ \\ =3\sqrt{2}-2\sqrt{3}+2\sqrt{3}-2 -3\sqrt{2}+3= -2 + 3 = \boxed{1 \in Z}[/tex]



[tex]b= \sqrt{11- \sqrt{96} } + \sqrt{14-4 \sqrt{6}} - \sqrt{4-2 \sqrt{3}}= \\ \\ = \sqrt{11- 4\sqrt{6} } + \sqrt{14-4 \sqrt{6}} - \sqrt{4-2 \sqrt{3}}= \\ \\ = \sqrt{8+3- 2\cdot2\sqrt{2}\cdot\sqrt{3} } + \sqrt{12+2- 2\cdot 2 \sqrt{3} \cdot \sqrt{2} } - \\ - \sqrt{3+1-2\cdot \sqrt{3} \cdot 1}= \\ \\ = \sqrt{8- 2\cdot2\sqrt{2}\cdot\sqrt{3} +3} + \sqrt{12 - 2\cdot 2 \sqrt{3} \cdot \sqrt{2} +2} - \\ - \sqrt{3-2\cdot \sqrt{3} \cdot 1 +1}= [/tex]

[tex]= \sqrt{(2\sqrt{2})^2- 2\cdot2\sqrt{2}\cdot\sqrt{3} +(\sqrt{3})^2} + \\ + \sqrt{(2 \sqrt{3})^2 - 2\cdot 2 \sqrt{3} \cdot \sqrt{2} +(\sqrt{2} )^2} - \sqrt{(\sqrt{3})^2-2\cdot \sqrt{3} \cdot 1 +1^2}= \\ \\ = \sqrt{(2\sqrt{2}-\sqrt{3})^2} + \sqrt{(2 \sqrt{3}-\sqrt{2} )^2} - \sqrt{(\sqrt{3}-1)^2}= \\ \\ =2\sqrt{2}-\sqrt{3} + 2 \sqrt{3}-\sqrt{2} -(2 \sqrt{3}-\sqrt{2} )= \\ \\ =2\sqrt{2}-\sqrt{3} + 2 \sqrt{3}-\sqrt{2} -2 \sqrt{3}+\sqrt{2} =\boxed{2\sqrt{2}-\sqrt{3} \notin Z}[/tex]



[tex]c=3\sqrt{8-\sqrt{48}}+\sqrt{43-30\sqrt{2}}+\sqrt{25-4\sqrt{6}}=\\\\ =3\sqrt{8-\sqrt{4\cdot 12}}+\sqrt{43-30\sqrt{2}}+\sqrt{25-4\sqrt{6}}=\\ \\ = 3\sqrt{8- 2\sqrt{ 12} } + \sqrt{43 -30\sqrt{2} } + \sqrt{25- 4\sqrt{6} } = \\ \\ = 3\sqrt{6+2- 2\cdot \sqrt{6} \sqrt{ 2} } + \sqrt{25+18 -2\cdot 5\cdot 3\sqrt{2} } + \\ + \sqrt{24+1- 2\cdot 2\sqrt{6}\cdot 1 } = \\ \\ = 3\sqrt{6- 2\cdot \sqrt{6} \sqrt{ 2}+2} + \sqrt{25 -2\cdot 5\cdot 3\sqrt{2} +18} + \\ + \sqrt{24-2\cdot 2\sqrt{6}\cdot 1 +1} = [/tex]

[tex]= 3\sqrt{(\sqrt{6})^2- 2\cdot \sqrt{6} \sqrt{ 2}+(2)^2} + \sqrt{5^2 -2\cdot 5\cdot 3\sqrt{2} +(3\sqrt{2})^2 } + \\ + \sqrt{(2\sqrt{6})^2-2\cdot 2\sqrt{6}\cdot 1 +1^2} = \\ \\ = 3\sqrt{(\sqrt{6}-2)^2} + \sqrt{(5-3\sqrt{2})^2 } + \sqrt{(2\sqrt{6}-1)^2} = \\ \\ 3\sqrt{6}-6+5-3\sqrt{2}+2\sqrt{6}-1= \boxed{5\sqrt{6}-3\sqrt{2} -2 \notin Z } [/tex]