Respuesta:
Explicación paso a paso:
radicales dobles
[tex]\sqrt{A +\sqrt{B} }[/tex] = [tex]\sqrt{\frac{A+C}{2} }[/tex] + [tex]\sqrt{\frac{A-C}{2} }[/tex]
Donde C =[tex]\sqrt{A^{2}-B }[/tex]
Para [tex]\sqrt{2+\sqrt{3} }[/tex] =>
C = [tex]\sqrt{2^{2} -3} = \sqrt{4-3} = \sqrt{1}[/tex]
C = 1
[tex]\sqrt{2+\sqrt{3} } = \sqrt{\frac{2+1}{2} } + \sqrt{\frac{2-1}{2} }[/tex]
[tex]\sqrt{2+\sqrt{3} } = \sqrt{\frac{3}{2} } + \sqrt{\frac{1}{2} }[/tex]
Para [tex]\sqrt{2-\sqrt{3} }[/tex] =>
C = [tex]\sqrt{2^{2} -3} = \sqrt{4-3} = 1[/tex]
[tex]\sqrt{2-\sqrt{3} } = \sqrt{\frac{2+1}{2} } - \sqrt{\frac{2-1}{2} }[/tex]
[tex]\sqrt{2-\sqrt{3} } = \sqrt{\frac{3}{2} } - \sqrt{\frac{1}{2} }[/tex]
Reemplaza
[ [tex]\sqrt{2+\sqrt{3} } - \sqrt{2-\sqrt{3} }[/tex] ]
[tex]\sqrt{\frac{3}{2} } + \sqrt{\frac{1}{2} } -( \sqrt{\frac{3}{2} } - \sqrt{\frac{1}{2} })[/tex] = [tex]\sqrt{\frac{1}{2} }[/tex] + [tex]\sqrt{\frac{1}{2} }[/tex] = [tex]2\sqrt{\frac{1}{2} } = 2.\frac{1}{\sqrt{2} } = \frac{2}{\sqrt{2} }[/tex]
racionalizando
[tex]\frac{2}{\sqrt{2} } . \frac{\sqrt{2} }{ \sqrt{2} } = \frac{2\sqrt{2} }{(\sqrt{2} )^{2} } = \sqrt{2}[/tex]
