| Resolviendo |
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Para eliminar el negativo del exponente, se invierte la base: |
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$N=\left(\dfrac{1}{2}\right)^{-\left(\dfrac{1}{2}\right)^{\textstyle -1}}+\left(\dfrac{1}{3}\right)^{-\left(\dfrac{1}{3}\right)^{\textstyle -1}}+\left(\dfrac{1}{4}\right)^{-\left(\dfrac{1}{4}\right)^{\textstyle -1}}$ |
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$N=\left(\dfrac{1}{2}\right)^{-\left(\dfrac{2}{1}\right)^{\textstyle 1}}+\left(\dfrac{1}{3}\right)^{-\left(\dfrac{3}{1}\right)^{\textstyle 1}}+\left(\dfrac{1}{4}\right)^{-\left(\dfrac{4}{1}\right)^{\textstyle 1}}$ |
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$N=\left(\dfrac{1}{2}\right)^{-\left(\textstyle 2 \right)^{\textstyle 1}}+\left(\dfrac{1}{3}\right)^{-\left(\textstyle 3 \right)^{\textstyle 1}}+\left(\dfrac{1}{4}\right)^{-\left(\textstyle 4 \right)^{\textstyle 1}}$ |
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$N=\left(\dfrac{1}{2}\right)^{\textstyle -2}+\left(\dfrac{1}{3}\right)^{-\textstyle 3 }+\left(\dfrac{1}{4}\right)^{-\textstyle 4 }$ |
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Para eliminar el negativo del exponente, se invierte la base: |
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$N=\left(\dfrac{2}{1}\right)^{\textstyle 2}+\left(\dfrac{3}{1}\right)^{\textstyle 3 }+\left(\dfrac{4}{1}\right)^{\textstyle 4 }$ |
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$N=\left(2\right)^{\textstyle 2}+\left(3 \right)^{\textstyle 3 }+\left(4 \right)^{\textstyle 4 }$ |
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$N=4+27+256$ |
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$N=287$ |
| Respuesta: |
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La solución es la Alternativa A |
