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- https://espanol.libretexts.org/Matematicas/Aritmetica_y_Matematicas_Basicas/HiSet_Matematicas/16%3A_Propiedades_basicas_de_los_numeros_reales/16.06%3A_Reglas_de_Exponentes\(x^{n}=\underbrace{x \cdot x \cdot x \cdot \ldots \cdot x}_{n \text { factors of } x}\) x^ {2}\ cdot x^ {4} =\ underbrackets {x x} _ {2}\ cdot\ underbrackets {x x x} _ {4} =\ underbrackets {x x x x x...\(x^{n}=\underbrace{x \cdot x \cdot x \cdot \ldots \cdot x}_{n \text { factors of } x}\) x^ {2}\ cdot x^ {4} =\ underbrackets {x x} _ {2}\ cdot\ underbrackets {x x x} _ {4} =\ underbrackets {x x x x x x} _ {6} =x^ {6}\\ \ dfrac {x^ {5}} {x^ {2}} =\ dfrac {x x x x} {x x} =\ dfrac {\ cancel {(x x)} x x x} {\ cancelar {(x x)}} =x x x=x^ {3}. \ text {Observe que} 5-2=3 Así, por la segunda regla de exponentes,\(\dfrac{x^n}{x^n} = x^{n-n} = x^0\)
- https://espanol.libretexts.org/Matematicas/Algebra/Algebra_elemental_(Ellis_y_Burzynski)/02%3A_Propiedades_basicas_de_los_numeros_reales/2.06%3A_Reglas_de_Exponentes\(x^{n}=\underbrace{x \cdot x \cdot x \cdot \ldots \cdot x}_{n \text { factors of } x}\) x^ {2}\ cdot x^ {4} =\ underbrackets {x x} _ {2}\ cdot\ underbrackets {x x x} _ {4} =\ underbrackets {x x x x x...\(x^{n}=\underbrace{x \cdot x \cdot x \cdot \ldots \cdot x}_{n \text { factors of } x}\) x^ {2}\ cdot x^ {4} =\ underbrackets {x x} _ {2}\ cdot\ underbrackets {x x x} _ {4} =\ underbrackets {x x x x x x} _ {6} =x^ {6}\\ \ dfrac {x^ {5}} {x^ {2}} =\ dfrac {x x x x} {x x} =\ dfrac {\ cancel {(x x)} x x x} {\ cancelar {(x x)}} =x x x=x^ {3}. \ text {Observe que} 5-2=3 Así, por la segunda regla de exponentes,\(\dfrac{x^n}{x^n} = x^{n-n} = x^0\)
- https://espanol.libretexts.org/Matematicas/Aritmetica_y_Matematicas_Basicas/HiSet_Mathematicas_Saul_Lopez/16%3A_Propiedades_basicas_de_los_numeros_reales/16.06%3A_Reglas_de_Exponentes\(x^{n}=\underbrace{x \cdot x \cdot x \cdot \ldots \cdot x}_{n \text { factors of } x}\) x^ {2}\ cdot x^ {4} =\ underbrackets {x x} _ {2}\ cdot\ underbrackets {x x x} _ {4} =\ underbrackets {x x x x x...\(x^{n}=\underbrace{x \cdot x \cdot x \cdot \ldots \cdot x}_{n \text { factors of } x}\) x^ {2}\ cdot x^ {4} =\ underbrackets {x x} _ {2}\ cdot\ underbrackets {x x x} _ {4} =\ underbrackets {x x x x x x} _ {6} =x^ {6}\\ \ dfrac {x^ {5}} {x^ {2}} =\ dfrac {x x x x} {x x} =\ dfrac {\ cancel {(x x)} x x x} {\ cancelar {(x x)}} =x x x=x^ {3}. \ text {Observe que} 5-2=3 Así, por la segunda regla de exponentes,\(\dfrac{x^n}{x^n} = x^{n-n} = x^0\)
