Question

Matemática potência , ajudem!!!

Answer 1

Olá!

Antes de começarmos, vamos revisar algumas propriedades dos expoentes:

$\sf a^m = \begin{matrix} \underbrace{\sf a \cdot a \cdot a \cdot \, ... \, \cdot a} \\ {\sf m~vezes} \end{matrix}$

$\sf a^m \cdot a^n = a^{m + n}$

$\sf \dfrac{a^m}{a^n} = a^{m - n}$

$\sf a^{-n} = \dfrac{1}{a^n}$

$\sf a^0 = 1$

$\sf a^1 = a$

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$\displaystyle \sf \frac{(-5)^2 - (3)^2 + (\frac{2}{3})^0}{3^{-2}+ \frac{1}{5} + \frac{1}{2}}$

$\displaystyle \sf \frac{25 - 9 + 1}{\frac{1}{3^2} + \frac{1}{5} + \frac{1}{2}}$

$\displaystyle \sf \frac{17}{\frac{1}{9} + \frac{1}{5} + \frac{1}{2}}$

→ Vamos calcular o denominador:

$\displaystyle \sf \frac{17}{\color{Red} \frac{1}{9} + \frac{1}{5} + \frac{1}{2}}$

$\displaystyle \sf \frac{1}{9} + \frac{1}{5} + \frac{1}{2}$

$\displaystyle \sf \frac{10}{90} + \frac{18}{90} + \frac{45}{90} = \frac{10 + 18 + 45}{90}$

$\fbox{\fbox{ \sf \dfrac{73}{90} }}$

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$\displaystyle \sf \frac{17}{\frac{1}{9} + \frac{1}{5} + \frac{1}{2}} = \frac{17}{\frac{73}{90}}$

$\displaystyle \sf 17 \cdot \frac{90}{73} = \frac{1.530}{73}$

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$\fbox{\fbox{ \displaystyle \sf \therefore \frac{(-5)^2 - (3)^2 + (\frac{2}{3})^0}{3^{-2}+ \frac{1}{5} + \frac{1}{2}} = \color{Red} \frac{\color{Red} 1.530}{\color{Red} 73} }}$

Espero ter ajudado.

Abraços e bons estudos ;-)