Question

810÷3³-4׳√27-(2⁸-3⁵)+2²​

Answer 1

RESOLVER:

$\frac{810}{3^3}-4\sqrt[3]{27}-\left(2^6\left(-3\right)^5\right)+2^2$

Quitamos los paréntesis

$=\frac{810}{3^3}-4\sqrt[3]{27}-2^6\left(-3\right)^5+2^2$

$\frac{810}{3^3}:30$

$4\sqrt[3]{27}:12$

$2^6\left(-3\right)^5:-2^6\cdot \:3^5$

$=30-12-\left(-2^6\cdot \:3^5\right)+2^2$

$=30-12+2^6\cdot \:3^5+2^2$

$30-12=18$

$=2^6\cdot \:3^5+2^2+18$

$2^6\cdot \:3^5:15552$

$2^2:4$

$=15552+4+18$

$\boxed{=15574}$

$Buena\:suerte\:con\:tus\:tareas\::ProfeAndresFelipe :)$