Question

Simplificar la siguiente expresión.​

Answer 1

E = 1 → RESPUESTA

Para resolver este ejercicio vamos a operar cada miembro por separado y finalmente calcularemos E. Tenemos que:

$\sqrt{\sqrt{\dfrac{32^n+8}{2^{5n-3}+1}+\dfrac{16}{5}\sqrt[n]{\dfrac{90^n}{6^{2n}}}}}\\\\ = \sqrt[4]{\dfrac{2^{5n}+2^3}{2^{5n-3}+1}+\dfrac{16}{5}\cdot\dfrac{90}{36}} \\\\= \sqrt[4]{\dfrac{2^{3}(2^{5n-3}+1)}{2^{5n-3}+1}+2^3} \\\\= \sqrt[4]{2^3+2^3} \\\\= \sqrt[4]{16} \\\\= 2$

$\sqrt[n]{\dfrac{32^n+16^n}{8^n+4^n}}\\\\=\sqrt[n]{\dfrac{2^{5n}+2^{4n}}{2^{3n}+2^{2n}}} \\\\= \sqrt[n]{\dfrac{2^{4n}(2^n+1)}{2^{2n}(2^n+1)}} \\\\= \sqrt[n]{\dfrac{2^{4n}}{2^{2n}}}\\\\ =\sqrt[n]{2^{2n}} \\\\= 2^2 \\\\= 4$

$\sqrt[n+1]{\dfrac{2^{3n+2}}{8^{n+1}-2^{3n+2}}} \\\\= \sqrt[n+1]{\dfrac{2^{3(n+1)-1}}{2^{3(n+1)}-2^{3(n+1)-1}}}\\\\=\sqrt[n+1]{\dfrac{2^{3(n+1)}2^{-1}}{2^{3(n+1)}(1-2^{-1})}}\\\\=\sqrt[n+1]{\dfrac{2^{3(n+1)}\cdot0.5}{2^{3(n+1)}(0.5)}}\\\\ = \sqrt[n+1]{1} \\\\= 1$

$\left( \sqrt[n-1]{\dfrac{3^{1-n}+1}{3^{n-1}+1}}\right)^{-1} \\\\= \sqrt[1-n]{\dfrac{3^{1-n}+1}{3^{n-1}+1}} \\\\= \sqrt[1-n]{\dfrac{3^{1-n}(1+3^{n-1})}{3^{n-1}+1}} \\\\= \sqrt[1-n]{3^{1-n}} \\\\= 3$

Finalmente queda:

$\boxed{E = \dfrac{2-4}{1-3} = \dfrac{-2}{-2} = 1}$