Resposta:
40/41
Explicação passo-a-passo:
[tex]\dfrac{3^2-3^{-2}}{3^2+3^{-2}}=\dfrac{3^2-\dfrac{1}{3^2}}{3^2+\dfrac{1}{3^2}}=\dfrac{\dfrac{3^2(3^2)-1}{3^2}}{\dfrac{3^2(3^2)+1}{3^2}}=[/tex][tex]=\dfrac{3^4-1}{3^2}\bigg(\dfrac{3^2}{3^4+1}\bigg)=\dfrac{3^4-1}{3^4+1}=\dfrac{81-1}{81+1}=\dfrac{80}{82}=\boxed{\dfrac{40}{41}}[/tex]
