Ejercicio 227 – Solucionario Baldor

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CAPITULO XXX

TEORIA DE LOS EXPONENTES

Hallar el valor de:

$\begin{matrix} {(a^{- 1})}^{2} & = a^{- 1 (2)} \\ = a^{- 2} \end{matrix}$
$\begin{matrix} {(a^{- 2} b^{- 1})}^{3} & = a^{- 2 (3)} b^{- 1 (3)} \\ = a^{- 6} b^{- 3} \end{matrix}$
$\begin{matrix} {(a^{\frac{3}{2}})}^{2} & = a^{\frac{3}{2} (2)} \\ = a^{3} \end{matrix}$
$\begin{matrix} {(x^{\frac{3}{4}})}^{3} & = x^{\frac{3}{4} (3)} \\ = x^{\frac{9}{4}} \end{matrix}$
$\begin{matrix} {(m^{\frac{3}{4}})}^{2} & = m^{\frac{3}{} (2)} \\ = m^{\frac{3}{2}} \end{matrix}$
$\begin{matrix} {(a^{- \frac{2}{3}})}^{3} & = a^{- \frac{2}{3} (3)} \\ = a^{- 2} \end{matrix}$
$\begin{matrix} {(x^{- 4} y^{\frac{1}{4}})}^{2} & = x^{- 4 (2)} y^{\frac{1}{} (2)} \\ = x^{- 8} y^{\frac{1}{2}} \end{matrix}$
$\begin{matrix} {(2 a^{\frac{1}{2}} b^{\frac{1}{3}})}^{2} & = 2^{2} a^{\frac{1}{2} (2)} b^{\frac{1}{3} (2)} \\ = 4 a b^{\frac{2}{3}} \end{matrix}$
$\begin{matrix} {(a^{- 3} b^{- 1})}^{4} & = a^{- 3 (4)} b^{- 1 (4)} \\ = a^{- 12} b^{- 4} \end{matrix}$
$\begin{matrix} {(x^{\frac{2}{3}} y^{- \frac{1}{2}})}^{6} & = x^{\frac{2}{3} ()} y^{- \frac{1}{2} ()} \\ = x^{4} y^{- 3} \end{matrix}$
$\begin{matrix} {(3 a^{\frac{2}{5}} b^{- 3})}^{5} & = 3^{5} a^{\frac{2}{5} (5)} b^{- 3 (5)} \\ = 243 a^{2} b^{- 15} \end{matrix}$
$\begin{matrix} {(2 m^{- \frac{1}{2}} n^{- \frac{1}{3}})}^{3} & = 2^{3} m^{- \frac{1}{2} (3)} n^{- \frac{1}{3} (3)} \\ = 8 m^{- \frac{3}{2}} n^{- 1} \end{matrix}$