Problema
(Indicado a partir do 9º ano do E. F.)
Considere os números
[tex]\qquad \alpha=\sqrt{13}+\sqrt{10+2\sqrt{13}}[/tex]
e
[tex]\qquad \beta=\sqrt{5+2\sqrt{3}}+\sqrt{18-2\sqrt{3}+2\sqrt{65-26\sqrt{3}}}[/tex] .
Mostre que [tex]\alpha=\beta[/tex].
Solução
- Inicialmente, observe que
[tex]\qquad\qquad\beta=\sqrt{5+2\sqrt{3}}+\sqrt{18-2\sqrt{3}+2\sqrt{65-26\sqrt{3}}}[/tex]
[tex]\qquad\qquad\beta=\sqrt{5+2\sqrt{3}}+\sqrt{18-2\sqrt{3}+2\sqrt{13}\sqrt{5-2\sqrt{3}}}[/tex]
[tex]\qquad\qquad\beta=\sqrt{5+2\sqrt{3}}+\sqrt{\left(\sqrt{13}+\sqrt{5-2\sqrt{3}}\right)^2}[/tex]
[tex]\qquad\qquad\boxed{\beta=\sqrt{5+2\sqrt{3}}+\sqrt{13}+\sqrt{5-2\sqrt{3}}}[/tex].
- Agora, seja [tex]\boxed{x=\sqrt{5+2\sqrt{3}}+\sqrt{5-2\sqrt{3}}}[/tex].
Assim:
[tex]\qquad\qquad x^2=5+2\sqrt{3}+2\sqrt{5+2\sqrt{3}}\sqrt{5-2\sqrt{3}}+5-2\sqrt{3}[/tex][tex]\qquad\qquad x^2=10+2\sqrt{25-12} [/tex]
[tex]\qquad\qquad x^2=10+2\sqrt{13} [/tex]
[tex]\qquad\qquad x=\pm\sqrt{10+2\sqrt{13}}. [/tex]
Como [tex]x\gt 0[/tex], então [tex] \boxed{x=\sqrt{10+2\sqrt{13}}}[/tex].
Portanto, pelo exposto,
[tex]\qquad \qquad \alpha=\sqrt{13}+\sqrt{10+2\sqrt{13}}[/tex]
[tex]\qquad \qquad \alpha=\sqrt{13}+x[/tex]
[tex]\qquad \qquad \alpha=\sqrt{13}+\left[\sqrt{5+2\sqrt{3}}+\sqrt{5-2\sqrt{3}}\right][/tex]
[tex]\qquad \qquad \alpha=\sqrt{5+2\sqrt{3}}+\sqrt{13}+\sqrt{5-2\sqrt{3}}[/tex]
[tex]\qquad\qquad\alpha=\beta[/tex].
Solução elaborada pelos Moderadores do Blog.