Question

Como resolver um número elevado a uma raiz?

$3^{ \sqrt{2} }$

Resolva: $(\frac{ \sqrt{2} }{2} ) ( 2^{0,5} )^{-2} - ( \frac{6}{ 6^{ \sqrt{3} } } )^{ \sqrt{3} + 1 }$

Answer 1

$\dfrac{\sqrt{2}}{2}\cdot (2^{0,5})^{-2}-\left(\dfrac{6}{6^{\sqrt{3}}} \right )^{\sqrt{3}+1}\\ \\ \\ =\dfrac{\sqrt{2}}{2}\cdot (2^{1/2})^{-2}-(6^{1-\sqrt{3}})^{\sqrt{3}+1}\\ \\ \\ =\dfrac{\sqrt{2}}{2}\cdot 2^{\frac{1}{2}\,\cdot \,(-2)}-6^{(1-\sqrt{3})\,\cdot\, ({\sqrt{3}+1})}\\ \\ \\ =\dfrac{\sqrt{2}}{2}\cdot 2^{-1}-6^{\sqrt{3}+1-3-\sqrt{3}}\\ \\ \\ =\dfrac{\sqrt{2}}{2}\cdot \dfrac{1}{2}-6^{-2}\\ \\ \\ =\dfrac{\sqrt{2}}{4}-\dfrac{1}{36}\\ \\ \\ =\dfrac{9\sqrt{2}-1}{36}$

Answer 2

Olá,

$( \frac{ \sqrt{2} }{2})(2^{0,5})^{-2} = ( \frac{ \sqrt{2} }{2})( \sqrt{2} )^{-2} = (\frac{ \sqrt{2} }{2})( \frac{1}{ (\sqrt{2})^2}) = (\frac{ \sqrt{2} }{2})( \frac{1}{2}) = \frac{ \sqrt{2} }{4}$

$( \frac{6}{6^\sqrt{3} }})^{ \sqrt{3} +1} = (6^{1- \sqrt{3}})^{ \sqrt{3}+1} = 6^{(1- \sqrt{3})(1+ \sqrt{3})$

$6^{(1-\sqrt{3})(1+ \sqrt{3})} = 6^{(1+ \sqrt{3}- \sqrt{3}- \sqrt{3}^{2})} = 6^{(1- \sqrt{9})} = 6^{(1-3)} = 6^{-2} = \frac{1}{6^2} = \frac{1}{36}$

$Logo$

$\frac{ \sqrt{2} }{4} - \frac{1}{36} = \frac{9}{9}* \frac{ \sqrt{2} }{4} - \frac{1}{36} = \frac{9 \sqrt{2} }{36} - \frac{1}{36} = ( \frac{9 \sqrt{2}-1 }{36})$