Question

Assunto potência e radiação: Calcule

Answer 1

$\dfrac{a^{\,^{-1}\!\!\!\diagup\!\!_{9}}\cdot \left(a^{\,^{-1}\!\!\!\diagup\!\!_{3}}\right)^{2}}{a^{2}} : \left(\dfrac{-1}{a} \right )^{2}\\ \\ =\dfrac{a^{\,^{-1}\!\!\!\diagup\!\!_{9}}\cdot a^{\,^{-1}\!\!\!\diagup\!\!_{3}\,\cdot\,2}}{a^{2}} : \dfrac{1}{a^{2}}\\ \\ =\dfrac{a^{\,^{-1}\!\!\!\diagup\!\!_{9}}\cdot a^{\,^{-2}\!\!\!\diagup\!\!_{3}}}{a^{2}} : \dfrac{1}{a^{2}}\\ \\ =\dfrac{a^{\,^{-1}\!\!\!\diagup\!\!_{9}\,^{-2}\!\!\!\diagup\!\!_{3}}}{a^{2}}:\dfrac{1}{a^{2}}\\ \\ =\dfrac{a^{\,^{-1}\!\!\!\diagup\!\!_{9}}\cdot a^{\,^{-2}\!\!\!\diagup\!\!_{3}}}{a^{2}} : \dfrac{1}{a^{2}}\\ \\ =\dfrac{a^{\,^{-1}\!\!\!\diagup\!\!_{9}\,^{-6}\!\!\!\diagup\!\!_{9}}}{a^{2}}:\dfrac{1}{a^{2}}\\ \\ =\dfrac{a^{\,^{-1}\!\!\!\diagup\!\!_{9}}\cdot a^{\,^{-2}\!\!\!\diagup\!\!_{3}}}{a^{2}} : \dfrac{1}{a^{2}}\\ \\ =\dfrac{a^{\,^{-7}\!\!\!\diagup\!\!_{9}}}{a^{2}}:\dfrac{1}{a^{2}}$

$=\dfrac{a^{\,^{-7}\!\!\!\diagup\!\!_{9}}}{a^{2}}\times a^{2}\\ \\ =a^{\,^{-7}\!\!\!\diagup\!\!_{9}}$