Resposta:

[tex]\textsf{Leia abaixo}[/tex]

Explicação passo a passo:

[tex]\mathsf{\sqrt{(2 + \sqrt{3}})^r + \sqrt{(2 - \sqrt{3}})^r = 14}[/tex]

[tex]\mathsf{\sqrt{(2 - \sqrt{3}})^{-r} + \sqrt{(2 - \sqrt{3}})^r = 14}[/tex]

[tex]\mathsf{\dfrac{1}{\sqrt{(2 - \sqrt{3}})^{r}} + \sqrt{(2 - \sqrt{3}})^r = 14}[/tex]

[tex]\mathsf{1 + \sqrt{(2 - \sqrt{3}})^{r^2} = 14(\sqrt{2 - \sqrt{3}})^r}[/tex]

[tex]\mathsf{\sqrt{(2 - \sqrt{3}})^{r^2} - 14(\sqrt{2 - \sqrt{3}})^r + 1 = 0}[/tex]

[tex]\mathsf{y = \sqrt{(2 - \sqrt{3}})^{r}}[/tex]

[tex]\mathsf{y^2 - 14y + 1 = 0}[/tex]

[tex]\mathsf{\Delta = b^2 - 4.a.c}[/tex]

[tex]\mathsf{\Delta = (-14)^2 - 4.1.1}[/tex]

[tex]\mathsf{\Delta = 196 - 4}[/tex]

[tex]\mathsf{\Delta = 192}[/tex]

[tex]\mathsf{y = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{14 \pm \sqrt{192}}{2} \rightarrow \begin{cases}\mathsf{y' = \dfrac{14 + 8\sqrt{3}}{2} = 7 + 4\sqrt{3}}\\\\\mathsf{y'' = \dfrac{14 - 8\sqrt{3}}{2} = 7 - 4\sqrt{3}}\end{cases}}[/tex]

[tex]\mathsf{\sqrt{(2 - \sqrt{3}})^{r} = (7 - 4\sqrt{3})}[/tex]

[tex]\mathsf{(2 - \sqrt{3})^{\frac{r}{2}} = (7 - 4\sqrt{3})}[/tex]

[tex]\mathsf{log\:(2 - \sqrt{3})^{\frac{r}{2}} = log\:(7 - 4\sqrt{3})}[/tex]

[tex]\mathsf{\dfrac{r}{2}\:.\:log\:(2 - \sqrt{3}) = log\:(7 - 4\sqrt{3})}[/tex]

[tex]\mathsf{\dfrac{r}{2} = log_{2 - \sqrt{3}}\:(7 - 4\sqrt{3})}[/tex]

[tex]\mathsf{\dfrac{r}{2} = 2}[/tex]

[tex]\mathsf{r = 4}[/tex]

[tex]\mathsf{\sqrt{(2 - \sqrt{3}})^{r} = (7 + 4\sqrt{3})}[/tex]

[tex]\mathsf{(2 - \sqrt{3})^{\frac{r}{2}} = (7 + 4\sqrt{3})}[/tex]

[tex]\mathsf{log\:(2 - \sqrt{3})^{\frac{r}{2}} = log\:(7 + 4\sqrt{3})}[/tex]

[tex]\mathsf{\dfrac{r}{2}\:.\:log\:(2 - \sqrt{3}) = log\:(7 + 4\sqrt{3})}[/tex]

[tex]\mathsf{\dfrac{r}{2} = log_{2 - \sqrt{3}}\:(7 + 4\sqrt{3})}[/tex]

[tex]\mathsf{\dfrac{r}{2} = -2}[/tex]

[tex]\mathsf{r = -4}[/tex]

[tex]\boxed{\boxed{\mathsf{S = \{4;-4\}}}}[/tex]