Ensino Fundamental ⇒ (EPCAR-1986) Expressão Numérica Tópico resolvido

[erastóstones]

Depois de racionalizar [tex3]\sqrt{\frac{\sqrt{10}+1}{\sqrt{10}-1}}-\sqrt{\frac{\sqrt{10}-1}{\sqrt{10}+1}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\sqrt{\frac{\sqrt{10}+1}{\sqrt{10}-1}}-\sqrt{\frac{\sqrt{10}-1}{\sqrt{10}+1}}[/tex3]

obtém-se;

a) 1/3
c) 2/3
c) 5/3
d) -27
e) 1

[deOliveira]

[tex3]\sqrt{\frac{\sqrt{10}+1}{\sqrt{10}-1}}-\sqrt{\frac{\sqrt{10}-1}{\sqrt{10}+1}}=\\\sqrt{\frac{\sqrt{10}+1}{\sqrt{10}-1}\cdot\frac{\sqrt{10}+1}{\sqrt{10}+1}}-
\sqrt{\frac{\sqrt{10}-1}{\sqrt{10}+1}\cdot\frac{\sqrt{10}-1}{\sqrt{10}-1}}=\\
\sqrt{\frac{(\sqrt{10}+1)^2}{(\sqrt{10})^2-1^2}}-\sqrt{\frac{(\sqrt{10}-1)^2}{(\sqrt{10})^2-1^2}}=\\
\sqrt{\frac{(\sqrt{10}+1)^2}{9}}-\sqrt{\frac{(\sqrt{10}-1)^2}{9}}=\\
\frac{\sqrt{10}+1}{3}-\frac{\sqrt{10}-1}3=\\\frac{\sqrt{10}+1-\sqrt{10}+1}3=\\\boxed{\boxed{\frac23}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\sqrt{\frac{\sqrt{10}+1}{\sqrt{10}-1}}-\sqrt{\frac{\sqrt{10}-1}{\sqrt{10}+1}}=\\\sqrt{\frac{\sqrt{10}+1}{\sqrt{10}-1}\cdot\frac{\sqrt{10}+1}{\sqrt{10}+1}}-
\sqrt{\frac{\sqrt{10}-1}{\sqrt{10}+1}\cdot\frac{\sqrt{10}-1}{\sqrt{10}-1}}=\\
\sqrt{\frac{(\sqrt{10}+1)^2}{(\sqrt{10})^2-1^2}}-\sqrt{\frac{(\sqrt{10}-1)^2}{(\sqrt{10})^2-1^2}}=\\
\sqrt{\frac{(\sqrt{10}+1)^2}{9}}-\sqrt{\frac{(\sqrt{10}-1)^2}{9}}=\\
\frac{\sqrt{10}+1}{3}-\frac{\sqrt{10}-1}3=\\\frac{\sqrt{10}+1-\sqrt{10}+1}3=\\\boxed{\boxed{\frac23}}[/tex3]

Espero ter ajudado .