Lista de exercícios do ensino médio para impressão
Racionalizar o denominador da fração $\phantom{X}\dfrac{4}{\;2\,+\,\sqrt{3\,}\,+\,\sqrt{7\,}}\phantom{X}$

 



resposta:

DIFERENÇA DE QUADRADOS
$\,\boxed{\;a^2\,-\,b^2\,=\,(a\,+\,b)\,\centerdot\,(a\,-\,b)\,}$


Resolução:
Multiplicamos o numerador e o denominador da fração por $\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,$
$\,\dfrac{4}{\;2\,+\,\sqrt{3\,}\,+\,\sqrt{7\,}}\;=$ $\,\dfrac{4}{\;2\,+\,\sqrt{3\,}\,+\,\sqrt{7\,}}\,\centerdot\,\dfrac{\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,}{\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,}\,=$ $\,\dfrac{4(\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,)}{\;\left[\,2\,+\,\sqrt{3}\,+\,\sqrt{7}\,\right]\,\centerdot\,\left[\,\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,\right] \;}\,=$ $\,\dfrac{4(\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,)}{\;(2\,+\sqrt{3\,})^2\,-\,(\sqrt{7\,})^2\;}\,=$ $\dfrac{4(\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,)}{\;4\,+\,2\,\centerdot\,2\,\centerdot\,\sqrt{3\,}\,+\,3\,-\,7\;}\,=$ $\dfrac{4(\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,)}{\;4\sqrt{3\,}\;}\,=$ $\dfrac{\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,}{\;\sqrt{3\,}\;}\,=$ $\dfrac{\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,}{\;\sqrt{3\,}\;}\,\centerdot\,\dfrac{\;\sqrt{3\,}\;}{\;\sqrt{3\,}\;}\,=$ $\dfrac{\,2\sqrt{3\,}\,+\,(\sqrt{3\,})^2\,-\,\sqrt{3\,} \centerdot \sqrt{7\,}\,}{(\sqrt{3\,})^2}\,=$ $\,\dfrac{\,2\sqrt{3\,}\,+\,3\,-\,\sqrt{21\,}\,}{3}\;$
$\boxed{\,\dfrac{\,2\sqrt{3\,}\,+\,3\,-\,\sqrt{21\,}\,}{3}\;}$
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