Question

1.Si: $\frac{x}{y}$ + $\frac{y}{x}$ =2 ,calcula:
$\frac{8 x^{4}y + 3 x y^{4} }{ x^{5}+ 2 y^{5} }$

2.calcula el valor de :
S=$\sqrt[32]{1+3( 2^{2}+1)( 2^{4}+1)( 2^{8}+1)( 2^{16} +1)( 2^{32}+1 ) ( 2^{64}+1) }$

POR FAVOR ES PARA MAÑANA RESUELVAN LOS DOS :)

Answer 1

Te ayudo con el segundo:
Primero debes saber algunas cosas:
$1+x+x^2+x^3+...+x^{n-1}= \frac{x^n-1}{x-1} \\ \\ x^{2n}-1=(x^n+1)(x^n-1)$

Entonces en tu problema:
$1+2+2^2+2^3+...+2^{127}= \frac{2^{128}-1}{2-1}=2^{128}-1\\=(2^{64}+1)(2^{64}-1)\\=(2^{64}+1)(2^{32}+1)(2^{32}-1)\\=(2^{64}+1)(2^{32}+1)(2^{16}+1)(2^{16}-1)\\=(2^{64}+1)(2^{32}+1)(2^{16}+1)(2^8+1)(2^8-1)\\=(2^{64}+1)(2^{32}+1)(2^{16}+1)(2^8+1)(2^4+1)(2^4-1)\\=(2^{64}+1)(2^{32}+1)(2^{16}+1)(2^8+1)(2^4+1)(2^2+1)(2^2-1)\\=(2^{64}+1)(2^{32}+1)(2^{16}+1)(2^8+1)(2^4+1)(2^4-1)\\=(2^{64}+1)(2^{32}+1)(2^{16}+1)(2^8+1)(2^4+1)(2^2+1)(2+1)(2-1)$
$=(2^{64}+1)(2^{32}+1)(2^{16}+1)(2^8+1)(2^4+1)(2^2+1)(3)$

Entonces:
$\sqrt[32]{1+3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)(2^{64}+1)}\\= \sqrt[32]{1+2^{128}-1}= \sqrt[32]{2^{128}}=2^{128/32}=2^4=16$

Saludos!