Ensino Fundamental ⇒ Racionalização Tópico resolvido

[petras]

Encontrar o valor de N = [tex3]\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}[/tex3]

? (R:1)

[LucasPinafi]

[tex3]\omega = \frac{\sqrt{\sqrt 5 +2}+ \sqrt{\sqrt 5 -2}}{\sqrt{\sqrt 5 +1}} \\ \omega^2 = \left(\frac{\sqrt{\sqrt 5 +2}+\sqrt{\sqrt 5 - 2}}{\sqrt{\sqrt 5 +1}}\right)^2 = \frac{\sqrt 5 + 2 +2 \sqrt{(\sqrt 5 +2)(\sqrt 5 -2)}+ \sqrt 5 -2}{\sqrt 5 +1} \\ \omega^2 = \frac{2\sqrt 5 + 2 \sqrt{5-4}}{\sqrt 5 +1} = \frac{2(\sqrt 5 +1)}{\sqrt 5 +1}=2 \\ \omega >0 \therefore \omega = \sqrt 2 \\ \\ \sqrt{3- 2 \sqrt 2} =\sqrt{1 -2 \sqrt 2 +2}= \sqrt{(\sqrt 2 - 1)^2} = \sqrt 2 - 1 \\ I = \sqrt 2 - (\sqrt 2 - 1) = 1[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\omega = \frac{\sqrt{\sqrt 5 +2}+ \sqrt{\sqrt 5 -2}}{\sqrt{\sqrt 5 +1}} \\ \omega^2 = \left(\frac{\sqrt{\sqrt 5 +2}+\sqrt{\sqrt 5 - 2}}{\sqrt{\sqrt 5 +1}}\right)^2 = \frac{\sqrt 5 + 2 +2 \sqrt{(\sqrt 5 +2)(\sqrt 5 -2)}+ \sqrt 5 -2}{\sqrt 5 +1} \\ \omega^2 = \frac{2\sqrt 5 + 2 \sqrt{5-4}}{\sqrt 5 +1} = \frac{2(\sqrt 5 +1)}{\sqrt 5 +1}=2 \\ \omega >0 \therefore \omega = \sqrt 2 \\ \\ \sqrt{3- 2 \sqrt 2} =\sqrt{1 -2 \sqrt 2 +2}= \sqrt{(\sqrt 2 - 1)^2} = \sqrt 2 - 1 \\ I = \sqrt 2 - (\sqrt 2 - 1) = 1[/tex3]

[petras]

Grato Lucas1