Potencias | Raices | Práctica 0 ejercicio 3c

EJERCICIO 3:Calcule

(c) $81^{3/4}+\left(\dfrac{16}{49} \right)^{-\frac{1}{2}} +\left(\dfrac{64}{27} \right)^{2/3}
+ 32^{-4/5} + (2^{-6})^{2/3} + 3^{7/2}\cdot 3^{1/2}$


Solución:
(c) $81^{3/4}+\left(\dfrac{16}{49} \right)^{-\frac{1}{2}} +\left(\dfrac{64}{27} \right)^{2/3}
+ 32^{-4/5} + (2^{-6})^{2/3} + 3^{7/2}\cdot 3^{1/2} = $

$\sqrt[4]{81}^3+\left(\dfrac{49}{16} \right)^{\frac{1}{2}} +\left(\dfrac{64^{2/3}}{27^{2/3}} \right)
+ \left(\dfrac{1}{32} \right)^{4/5} + \left(\dfrac{1}{2} \right)^{6\cdot 2/3} + 3^{7/2 + 1/2} = $

$3^3+ \dfrac{\sqrt{49}}{\sqrt{16}} + \dfrac{\sqrt[3]{64}^2 }{\sqrt[3]{27}^2}
+ \dfrac{1}{\sqrt[5]{32}^4} + \left(\dfrac{1}{2} \right)^{4} + 3^{4} = $

$ 27 + \dfrac{7}{4} + \dfrac{4^2 }{3^2}
+ \dfrac{1}{2^4} + \dfrac{1}{2^{4}} + 81 = $

$ 27 + \dfrac{7}{4} + \dfrac{16 }{9}
+ \dfrac{1}{16} + \dfrac{1}{16} + 81 = $

$ 108 + \dfrac{28+1+1}{16} + \dfrac{16 }{9} = 108 + \dfrac{\cancel{30}^{15}}{\cancel{16}_8} + \dfrac{16 }{9} =$

$= 108 + \dfrac{15}{8} + \dfrac{16 }{9} = \dfrac{72\cdot 108 + 9\cdot 15 + 16\cdot 8 }{72} = \boxed{\dfrac{8039}{72} }$