Resposta:

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

Explicação passo a passo:

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

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

[tex]\mathsf{\dfrac{1}{\sqrt{(2 - \sqrt{3}})^{x}} + \sqrt{(2 - \sqrt{3}})^x = 4}[/tex]

[tex]\mathsf{1 + \sqrt{(2 - \sqrt{3}})^{x^2} = 4(\sqrt{2 - \sqrt{3}})^x}[/tex]

[tex]\mathsf{\sqrt{(2 - \sqrt{3}})^{x^2} - 4(\sqrt{2 - \sqrt{3}})^x + 1 = 0}[/tex]

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

[tex]\mathsf{y^2 - 4y + 1 = 0}[/tex]

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

[tex]\mathsf{\Delta = (-4)^2 - 4.1.1}[/tex]

[tex]\mathsf{\Delta = 16 - 4}[/tex]

[tex]\mathsf{\Delta = 12}[/tex]

[tex]\mathsf{y = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{4 \pm \sqrt{12}}{2} \rightarrow \begin{cases}\mathsf{y' = \dfrac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}}\\\\\mathsf{y'' = \dfrac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}}\end{cases}}[/tex]

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

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

[tex]\mathsf{\dfrac{x}{2} = 1}[/tex]

[tex]\mathsf{x = 2}[/tex]

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

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

[tex]\mathsf{\dfrac{x}{2} = -1}[/tex]

[tex]\mathsf{x = -2}[/tex]

[tex]\boxed{\boxed{\mathsf{S = \{2;-2\}}}}[/tex]