Question

me ajudem a resolver

Answer 1

$x= (2^2)^3= 2^6 y= (2^3)^2 = 2^6 z= 2^{3 ^{2} }= 2^9$
     $x.y.z= 2^6.2^6.2^9= 2^2^1$
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$4.( \frac{1}{2})^4+( \frac{1}{2}) ^{ \frac{1}{2} }=$

$4. \frac{1}{16} + \sqrt{ \frac{1}{2} }= \frac{1}{4}+ \frac{ \sqrt{2} }{2}= \frac{1+2 \sqrt{2} }{4} = \frac{1+ \sqrt{2} }{2}$
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Resolvendo separadamente:
 $16^{ \frac{3}{4} }= \sqrt[4]{16^3} = \sqrt[4]{(2^4)^3}= \sqrt[4]{2^1^2}=2^3= 8$

$8^{ \frac{1}{3} }= \sqrt[3]{8}= \sqrt[3]{2^3}=2$

Então fica:
$\frac{8}{2}: \frac{2^4}{8^2}= \frac{8}{2}* \frac{16}{64}= \frac{8}{8} = 1$