Respostas

2014-08-21T10:34:52-03:00
E ae meu brother,

use as propriedades da exponenciação:

\left[\left(8^{ \tfrac{1}{2}\right)^4\right]^{ \tfrac{1}{6}}+16^{ \tfrac{1}{4}}-27^{ \tfrac{2}{3}}\\\\\\
\left[\left(2^{ \tfrac{3}{2}\right)^4\right]^{ \tfrac{1}{6}}+(2^4)^{ \tfrac{1}{4}}-(3^3)^{ \tfrac{2}{3}}\\\\\\
\left(2^{ \tfrac{12}{2}\right)^{ \tfrac{1}{6}}+2^1-3^1\\\\\\
\left(2^6\right)^{ \tfrac{1}{6}}+2-3\\\\\\
2^1+2-3\\\\\\
2+2-3\\\\\\
1

Portanto,

\large\boxed{\boxed{\boxed{\left[\left(8^{ \tfrac{1}{2}\right)^4\right]^{ \tfrac{1}{6}}+16^{ \tfrac{1}{4}}-27^{ \tfrac{2}{3}}=1}}}}.\\.

>>>>>>>>>>>>>>>>>>>>>>>>>>>>

 \dfrac{0,005\cdot0,75\cdot(0,5)^{-2}}{0,125\cdot(0,25)^{-1}}= \dfrac{ \dfrac{5}{1.000}\cdot \dfrac{75}{100}  \cdot \dfrac{50}{100}^{-2} }{ \dfrac{125}{1.000}\cdot \dfrac{25}{100}^{-1}  }\\\\\\
 \dfrac{0,005\cdot0,75\cdot(0,5)^{-2}}{0,125\cdot(0,25)^{-1}}= \dfrac{ \dfrac{5\div5}{1.000\div5} \cdot \dfrac{75\div25}{100\div25}\cdot \dfrac{50\div50}{100\div50}^{-2}  }{ \dfrac{125\div125}{1.000\div125}\cdot \dfrac{25\div25}{100\div25} ^{-1} }


 \dfrac{0,005\cdot0,75\cdot(0,5)^{-2}}{0,125\cdot(0,25)^{-1}}= \dfrac{ \dfrac{1}{200}\cdot \dfrac{3}{4}\cdot \dfrac{1}{2}^{-2}   }{ \dfrac{1}{8}\cdot \dfrac{1}{4}^{-1}  }\\\\\\
 \dfrac{0,005\cdot0,75\cdot(0,5)^{-2}}{0,125\cdot(0,25)^{-1}}= \dfrac{ \dfrac{3}{800}\cdot(2^{-1})^{-2} }{ \dfrac{1}{8}\cdot(2^{-2})^{-1} }

 \dfrac{0,005\cdot0,75\cdot(0,5)^{-2}}{0,125\cdot(0,25)^{-1}}= \dfrac{ \dfrac{3}{800}\cdot\not2^2 }{ \dfrac{1}{8} \cdot\not2^2}\\\\\\
 \dfrac{0,005\cdot0,75\cdot(0,5)^{-2}}{0,125\cdot(0,25)^{-1}}= \dfrac{3}{800}\div \dfrac{1}{8}

 \dfrac{0,005\cdot0,75\cdot(0,5)^{-2}}{0,125\cdot(0,25)^{-1}}= \dfrac{3}{800}\cdot \dfrac{8}{1}\\\\\\
 \Large\boxed{\boxed{\boxed{\dfrac{0,005\cdot0,75\cdot(0,5)^{-2}}{0,125\cdot(0,25)^{-1}}= \dfrac{24\div8}{800\div8}= \dfrac{3}{100}}}}.\\.

E portanto alternativa E.

Tenha ótimos estudos ;D
1 5 1