Question

ayidaa .................

Answer 1

$1.- Reducir :$$\frac{\left(6+\frac{1}{2}\right)^4\cdot \left(2-\frac{1}{6}\right)^2}{\left(2+\frac{1}{6}\right)^4\cdot \left(6-\frac{1}{2}\right)^2}$

$* \left(2+\frac{1}{6}\right)^4=\left(\frac{13}{6}\right)^4=\frac{13^4}{6^4}$

$* \left(6-\frac{1}{2}\right)^2=\left(\frac{11}{2}\right)^2=\frac{11^2}{2^2}$

$*\left(6+\frac{1}{2}\right)^4=\left(\frac{13}{2}\right)^4=\frac{13^4}{2^4}$

$*\left(2-\frac{1}{6}\right)^2= \left(\frac{11}{6}\right)^2=\frac{11^2}{6^2}$

$\frac{\frac{13^4}{2^4}\cdot \frac{11^2}{6^2}}{\frac{13^4}{6^4}\cdot \frac{11^2}{2^2}}$

$\frac{\frac{3455881}{576}}{\frac{3455881}{5184}}$

$\frac{5184}{576}$$=9$


$2.- Reducir:$$P = 64^{-9^{-4^{-2^{-1}}}}$

$4^{-2^{-1}}=\frac{1}{2}$

$9^{-\frac{1}{2}}$

$9^{-\frac{1}{2}}=\frac{1}{\sqrt{9}}=\frac{1}{3}$

$P= 64^{-\frac{1}{3}}$

$64^{-\frac{1}{3}}=\frac{1}{\sqrt[3]{64}}=\frac{1}{4}$


$Simplificar:$$\frac{2^{m+1}\cdot 4^{m+2n}}{8^{m-1}\cdot 16^{^{n+1}}}$

$\mathrm{Factorizar}\:8^{m-1}:\quad 2^{3\left(m-1\right)}$

$\mathrm{Factorizar}\:4^{m+2n}:\quad 2^{2\left(m+2n\right)}$

$\mathrm{Factorizar}\:16^{n+1}:\quad 2^{4\left(n+1\right)}$

$=\frac{2^{2\left(m+2n\right)}\cdot \:2^{m+1}}{2^{3\left(m-1\right)}\cdot \:2^{4\left(n+1\right)}}$

$\mathrm{Simplificar}$

$\frac{2^{m+1+2\left(m+2n\right)}}{2^{3\left(m-1\right)+4\left(n+1\right)}}$

$\mathrm{Eliminar\:los\:terminos\:comunes:}\:2^{m+1+2\left(m+2n\right)}$

$1$