Ensino Fundamental ⇒ (OM-Sergipana-1999) Expressão Numérica Tópico resolvido

[erastóstones]

Simplificar [tex3]2-\(2-\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2-\sqrt{3}}}\)^2[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]2-\(2-\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2-\sqrt{3}}}\)^2[/tex3]

[deOliveira]

[tex3]2-\(2-\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2-\sqrt{3}}}\)^2=\\
2-\left(2-\frac{\sqrt{2+\sqrt3}}{\sqrt{2-\sqrt3}}\cdot\frac{\sqrt{2+\sqrt3}}{\sqrt{2+\sqrt3}}\right)^2=\\
2-\left(2-\frac{(\sqrt{2+\sqrt3})^2}{\sqrt{4-3}}\right)^2=\\
2-\left(2-(2+\sqrt3)\right)^2=\\2-(2-2-\sqrt3)^2=\\2-(-\sqrt3)^2=\\2-3=\\-1[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]2-\(2-\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2-\sqrt{3}}}\)^2=\\
2-\left(2-\frac{\sqrt{2+\sqrt3}}{\sqrt{2-\sqrt3}}\cdot\frac{\sqrt{2+\sqrt3}}{\sqrt{2+\sqrt3}}\right)^2=\\
2-\left(2-\frac{(\sqrt{2+\sqrt3})^2}{\sqrt{4-3}}\right)^2=\\
2-\left(2-(2+\sqrt3)\right)^2=\\2-(2-2-\sqrt3)^2=\\2-(-\sqrt3)^2=\\2-3=\\-1[/tex3]

Espero ter ajudado .