Question

determine as potências​

Answer 1

Olá!

$a) {16}^{ \frac{1}{2} } = ({2^{4} )}^{ \frac{1}{2} } = {2}^{ \frac{4 \times 1}{2} } = {2}^{ \frac{4}{2} } = {2}^{2} = 4 \\ \\ b) {81}^{ \frac{1}{4} } = ( {3}^{4} )^{ \frac{1}{4} } = {3}^{ \frac{4 \times 1}{4} } = 3 \\ \\ c)({0.5)}^{ - 1} = \frac{1}{2}^{ - 1} = {2}^{1} = 2 \\ \\ d) {3}^{ - 5} = ({ \frac{1}{3} })^{5} = \frac{ {1}^{5} }{ {3}^{5} } = \frac{1}{243} \\ \\ e) {( - 5)}^{ - 3} = - \frac{1}{5}^{3} = - \frac{ {1}^{3} }{ {5}^{3} } = - \frac{1}{125} \\ \\ f) {64}^{ \frac{1}{2} } = ( {2}^{6})^{ \frac{1}{2} } = {2}^{ \frac{ 6 \times 1}{2} } = {2}^{ \frac{6}{2} } = {2}^{3} = 8 \\ \\ g) ({ \frac{3}{5} })^{ - 2} = ( \frac{5}{3} )^{2} = \frac{ {5}^{2} }{ {3}^{2} } ) = \frac{25}{9} \\ \\ h)(0.25)^{ - 2} = ( \frac{25}{100} )^{ - 2} = (\frac{1}{4} )^{ - 2} = {4}^{2} = 16 \\ \\ i)( \frac{3}{4} )^{ - 2} =( { \frac{4}{3} })^{2} = \frac{ {4}^{2} }{ {3}^{2} } = \frac{16}{9} \\ \\ j)(0.1)^{ - 2} = ( { \frac{1}{10} })^{ - 2} = {10}^{2} = 100 \\ \\ k)( - 15) ^{2} = {15}^{2} = 225 \\ \\ l) {1}^{0.2} = 1 \\ \\ m) {27}^{ \frac{2}{3} } = ( {3}^{3} ) ^{ \frac{2}{3} } = {3}^{ \frac{3 \times 2}{3} } = 3^{ \frac{6}{3} } = {3}^{2} = 9 \\ \\ n)32^{ \frac{1}{5} } = ( {2}^{5} )^{ \frac{1}{5} } = {2}^{ \frac{5 \times 1}{5} } = {2}^{ \frac{5}{5} } = {2}^{1} = 2 \\ \\ o)9^{ - \frac{1}{2} } = \frac{1}{9 \frac{1}{2} } = \frac{1}{ ({3}^{2})^{ \frac{1}{2} } } = \frac{1}{ {3}^{ \frac{2 \times 1}{2} } } = \frac{1}{ {3}^{ \frac{2}{2} } } = \frac{1}{ {3}^{1} } = \frac{1}{3} \\ \\ p) { \frac{3}{4} }^{0} = 1$

Espero ter ajudado. Bons estudos!!!