Question

operaciones combinadas de potencia radicación
procedimiento porfas​

Answer 1

Verified answer

Respuesta:

Pregunta 1

• [tex] {( {5}^{3} \times {2}^{2} \times {4}^{3} ) }^{2} = {5}^{3 \times 2} \times {2}^{2 \times 2} \times { {(2}^{2}) }^{3 \times 2} \\ {5}^{6} \times {2}^{4 + 12} = {5}^{6} \times {2}^{16} [/tex]
• [tex] {3}^{2} \times {3}^{3} ( \frac{ {3}^{3} \times {3}^{4} }{ {3}^{4} \times {3}^{2} } ) = {3}^{2 + 3} ( \frac{ {3}^{3 + 4} }{ {3}^{4 + 2} } ) = {3}^{5} ({3}^{7 - 6} ) = {3}^{5} \times {3}^{1} = {3}^{6} [/tex]
• [tex]( \frac{3 \times {2}^{3} }{ {2}^{2} \times {3}^{2} } )^{2} = {( \frac{ {2}^{3 - 2} }{ {3}^{2 - 1} }) }^{2} = ( \frac{2}{3} )^{2} = \frac{4}{9} [/tex]

Pregunta 2

1. [tex] { ({5}^{2}) }^{3} = {(5)}^{x} \\ {5}^{2 \times 3} = {5}^{x} \\ 6 = x[/tex]
2. [tex] {( {( \frac{3}{7}) }^{x}) }^{5} = {( \frac{3}{7}) }^{10} \\ { \frac{3}{7} }^{5x} = { \frac{3}{7} }^{10} \\ 5x = 10 \\ x = \frac{10}{5} \\ x = 2[/tex]
3. [tex] {( \frac{4}{5}) }^{x} = \frac{256}{625} \\ { (\frac{4}{5}) }^{x} = \frac{ {4}^{4} }{ {5}^{4} } \\ {( \frac{4}{5} )}^{x} = {( \frac{4}{5} )}^{4} \\ x = 4[/tex]

Pregunta 3

[tex] \sqrt[3]{ \frac{81}{3} } \times {3}^{4} \times \frac{1}{ {3}^{4} } \\ \sqrt[3]{( \frac{81}{3} )} \times 1 \\ \sqrt[3]{27} = 3[/tex]

[tex] \frac{ {( \sqrt{49} \times \sqrt{49} \times \sqrt{49} \times \sqrt{49}) }^{2} }{ {( { \sqrt{49} }^{2} )}^{3} } = \frac{ {( \sqrt{ {49}^{4} }) }^{2} }{ {\sqrt{49}}^{6} } = { \sqrt{49} }^{8 - 6} = { \sqrt{49} }^{2} = 49[/tex]

[tex] \frac{ {10}^{2} \times {10}^{3} \times \sqrt[3]{1000} \times {10}^{4} }{ {( \sqrt[4]{10000} \times \sqrt[3]{1000} ) }^{2} } = \frac{ {10}^{2 + 3 + 4 + 1} }{ {( {10}^{2}) }^{2} } = \frac{ {10}^{10} }{ {10}^{4} } = {10}^{10 - 4} = {10}^{6} [/tex]