Question

Como faz...
N^-2 para a expressão
N= 2^-1 + 2^ -1/2

Answer 1

$\begin{array}{l}\textsf{Primeiro devemos resolver N}\\\\\\ \mathsf{N=2^{-1}+~2^{-\frac{1}{2}}}\\\\\\\mathsf{=\dfrac{1}{2^1}~+~\dfrac{1}{2^{\frac{1}{2}}}}\\\\\\\mathsf{=\dfrac{1}{2}~+~\dfrac{1}{\sqrt{2}}}\\\\\\\mathsf{=\dfrac{1}{2}~+~\begin{pmatrix} \dfrac{1}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}} \end{pmatrix}}\\\\\\\mathsf{=\dfrac{1}{2}~+~\dfrac{\sqrt{2}}{2}}\\\\\\\mathsf{\large\fbox{ N=\dfrac{\sqrt{2}+1}{2} }}\end{array}$


$\begin{array}{l}\mathsf{Agora~devemos~resolver~N^{-2}:}\\\\\\\mathsf{\begin{pmatrix} \dfrac{\sqrt{2}+1}{2} \end{pmatrix}^{-2}}\\\\\\\mathsf{=\dfrac{1}{\begin{pmatrix} \dfrac{\sqrt{2}+1}{2} \end{pmatrix}^{2}}}\\\\\\=\mathsf{\dfrac{1}{\dfrac{(\sqrt{2})^2+2\sqrt{2}+1^2}{2^2}}}\\\\\\\mathsf{=\dfrac{1}{\dfrac{2+2\sqrt{2}+1}{4}}}\\\\\\\mathsf{=\dfrac{1}{\dfrac{3+2\sqrt{2}}{4}}}\end{array}$


$\begin{array}{l}\mathsf{=1\cdot\dfrac{4}{3+2 \sqrt{2}}}\\\\\\\mathsf{=\dfrac{4}{3+2\sqrt{2}}}\\\\\\\mathsf{=\dfrac{4}{(3+2\sqrt{2})}\cdot\dfrac{(3-2\sqrt{2})}{(3-2\sqrt{2})}}\\\\\\\mathsf{=\dfrac{12-8\sqrt{2}}{3^2-(2\sqrt{2})^2}}\\\\\\\mathsf{=\dfrac{12-8\sqrt{2}}{9-(4\cdot(\sqrt{2})^2)}}\\\\\\\mathsf{=\dfrac{12-8\sqrt{2}}{9-(4\cdot 2)}}\\\\\\\mathsf{=\dfrac{12-8\sqrt{2}}{9-8}}\end{array}$


$\begin{array}{l}\mathsf{=\dfrac{12-8\sqrt{2}}{1}}\\\\\\\large\fbox{ N^{-2}=12-8\sqrt{2} }\end{array}$


$\large\textsf{Obs: Se ficar com d\'uvidas revise exponencia\c{c}\~ao, radicia\c{c}\~ao}\\\textsf{e produtos not\'aveis.}$