Question

potencian alludeme???​

Answer 1

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

${1}^{10} = 1$

${0}^{100} = 0$

${( - 1)}^{9} = - 1$

${( - 1)}^{10} = 1$

${( - 2894320)}^{0} = 1$

${2}^{1} + {3}^{ - 2} = 2 + \frac{1}{ {3}^{2} } = 2 + \frac{1}{9} = \frac{19}{9}$

${( {2}^{ - 1} + {4}^{ - 1)} }^{ - 1} = {((1 + {2}^{ - 1}) \times {2}^{ - 1} )}^{ - 1} = {((1 + \frac{1}{2}) \times {2}^{ - 1} ) }^{ - 1} = {( \frac{3}{2} \times {2}^{ - 1} ) }^{ - 1} = {( \frac{3}{ {2}^{2} }) }^{ - 1} = \frac{ {2}^{2} }{3} = \frac{4}{3}$