• Matéria: Matemática
  • Autor: astro3347
  • Perguntado 4 anos atrás


 \frac{1}{ \sqrt{6 +  \sqrt{5} } }

Respostas

respondido por: elizeugatao
0

\displaystyle \frac{1}{\sqrt{6+\sqrt5}} \\\\\\ \underline{\text{racionalizando}}: \\\\ \frac{1}{(\sqrt{6+\sqrt5})}.\frac{\sqrt{6+\sqrt5}}{(\sqrt{6+\sqrt5})} \\\\\\ \frac{\sqrt{6+\sqrt5}}{(\sqrt{6+\sqrt5})^2} \\\\\\ \frac{\sqrt{6+\sqrt5}}{6+\sqrt5} \\\\ \underline{\text {racionalizando novamente}} :  \\\\ \frac{\sqrt{6+\sqrt5}}{(6+\sqrt5)}.\frac{6-\sqr5}{(6-\sqrt5)} \\\\\\ \frac{(6-\sqrt5).\sqrt{6+\sqrt5}}{6^2-(\sqrt5)^2} \\\\ \frac{(6-\sqrt5).\sqrt{6+\sqrt5}}{36-5} \to \frac{(6-\sqrt5).\sqrt{6+\sqrt5}}{31}

Fazendo a distributiva, temos :

\displaystyle \frac{6\sqrt{6+\sqrt5}-\sqrt5.\sqrt{6+\sqrt5}}{31} \\\\\\ \frac{6\sqrt{6+\sqrt5} - \sqrt{5(6+\sqrt5)}}{31}  \\\\\\ \frac{6\sqrt{6+\sqrt5} -\sqrt{30+5\sqrt{5}}}{31}

Portanto, temos :

\huge\boxed{\frac{(6-\sqrt5).\sqrt{6+\sqrt5}}{31}\ }\checkmark \\\\\\\\ \text{ou} \\\\\\\huge\boxed{\displaystyle \frac{6\sqrt{6+\sqrt5} -\sqrt{30+5\sqrt{5}}}{31}} \checkmark

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