Question

Resolver aplicando propiedades de la potencion,ayudaa!!
solo ejercicio C y D

Answer 1

C. $\left[\left(2-\frac{8}{5}\right)^2+\left(\frac{5}{8}-\frac{3}{4}\right)-\left(\frac{6}{5}.\:\frac{1}{3}\right)^4.\left(\frac{15}{2}\right)^3\right]:\left(5-\frac{6}{5}\right)$

$\frac{\left(\left(2-\frac{8}{5}\right)^2+\left(\frac{5}{8}-\frac{3}{4}\right)-\left(\frac{6}{5}\cdot \frac{1}{3}\right)^4\left(\frac{15}{2}\right)^3\right)}{\left(5-\frac{6}{5}\right)}$

$=\frac{\left(2-\frac{8}{5}\right)^2+\frac{5}{8}-\frac{3}{4}-\left(\frac{6}{5}\cdot \frac{1}{3}\right)^4\left(\frac{15}{2}\right)^3}{5-\frac{6}{5}}$

$5-\frac{6}{5}=\frac{19}{5}$

$=\frac{\left(-\frac{8}{5}+2\right)^2+\frac{5}{8}-\frac{3}{4}-\left(\frac{15}{2}\right)^3\left(\frac{6}{5}\cdot \frac{1}{3}\right)^4}{\frac{19}{5}}$

$\left(2-\frac{8}{5}\right)^2=\frac{2^2}{5^2}$

$\left(\frac{6}{5}\cdot \frac{1}{3}\right)^4\left(\frac{15}{2}\right)^3=\frac{54}{5}$

$=\frac{\frac{2^2}{5^2}+\frac{5}{8}-\frac{3}{4}-\frac{54}{5}}{\frac{19}{5}}$

$\frac{2^2}{5^2}=\frac{4}{25}$$=\frac{\frac{4}{25}+\frac{5}{8}-\frac{3}{4}-\frac{54}{5}}{\frac{19}{5}}$

$=\frac{5\left(\frac{5}{8}+\frac{4}{25}-\frac{3}{4}-\frac{54}{5}\right)}{19}$

$\frac{4}{25}+\frac{5}{8}-\frac{3}{4}-\frac{54}{5}=-\frac{2153}{200}$

$=\frac{5\left(-\frac{2153}{200}\right)}{19}$

$=\frac{-\frac{2153}{200}\cdot \:5}{19}$

$=-\frac{5\cdot \frac{2153}{200}}{19}$

$\frac{2153}{200}\cdot \:5\:=\frac{2153}{40}$

$=-\frac{\frac{2153}{40}}{19}$

$=-\frac{2153}{760}$

D. $\frac{\left(2-\frac{1}{5}\right)^2}{\left(3-\frac{2}{9}\right)^{-1}}:\frac{\left(\frac{6}{7}.\:\frac{5}{4}-\frac{2}{7}:\:\frac{1}{2}\right)^3}{\left(\frac{1}{2}-\frac{1}{3}.\:\frac{1}{4}:\:\frac{1}{5}\right)}-\frac{36}{7}$

$\frac{\frac{\left(2-\frac{1}{5}\right)^2}{\left(3-\frac{2}{9}\right)^{-1}}}{\frac{\left(\frac{6}{7}\cdot \frac{5}{4}-\frac{\frac{2}{7}}{\frac{1}{2}}\right)^3}{\frac{1}{2}-\frac{1}{3}\cdot \frac{\frac{1}{4}}{\frac{1}{5}}}}-\frac{36}{7}$

$\frac{\frac{\left(2-\frac{1}{5}\right)^2}{\left(3-\frac{2}{9}\right)^{-1}}}{\frac{\left(\frac{6}{7}\cdot \frac{5}{4}-\frac{\frac{2}{7}}{\frac{1}{2}}\right)^3}{\frac{1}{2}-\frac{1}{3}\cdot \frac{\frac{1}{4}}{\frac{1}{5}}}}=6$

$=6-\frac{36}{7}$

$=\frac{6\cdot \:7}{7}-\frac{36}{7}$

$=\frac{6\cdot \:7-36}{7}$

$6\cdot \:7-36=6$

$=\frac{6}{7}$

¡Espero haberte ayudado Biebsxshawn!