Question

para qual valor de p é verdadeiro a igualdade 3.2²-(3-²)-¹+0,2-³=8/P²?

Answer 1

Geslaine,
Vamos passo-a-passo

$3.2^2-(3^{-2} ) ^{-1} +(0,2) ^{-3} = \frac{8}{P^2} \\ \\ 12- (\frac{1}{9} ) ^{-1} +( \frac{2}{10} ) ^{-3} = \frac{8}{P^2} \\ \\ 12 - 9+5^3= \frac{8}{P^2} \\ \\ 128= \frac{8}{P^2} \\ \\ P^2= \frac{8}{128} \\ \\ P^2= \frac{1}{16} \\ \\ P= \sqrt{ \frac{1}{16} } \\ \\ P=+/- \frac{1}{4}$

Answer 2

3×2^2-(3^-2)^-1+(0,2)^-3
12-(1/9)^-1+(2/10^-3=8/P2
12-9+5^3=8/P^2
P2=1/16
P=raiz quadrada de 1/16
P+/-1/4