Resposta:

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

Explicação passo a passo:

[tex]\Large \boxed{\sf x^2 + \dfrac{1}{x} = 2}[/tex]

[tex]\Large \boxed{\sf x^3 + 1 = 2x}[/tex]

[tex]\Large \boxed{\sf x^3 - 2x + 1 = 0}[/tex]

[tex]\Large \boxed{\sf x_1 = 1}[/tex]

[tex]\Large \boxed{\sf \dfrac{x^3 - 2x + 1}{x - 1} = x^2 + x - 1}[/tex]

[tex]\Large \boxed{\sf x^2 + x - 1 = 0}[/tex]

[tex]\Large \boxed{\sf a = 1 \leftrightarrow b = 1 \leftrightarrow c = -1}[/tex]

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

[tex]\Large \boxed{\sf \Delta = 1^2 - 4.1.(-1)}[/tex]

[tex]\Large \boxed{\sf \Delta = 1 + 4}[/tex]

[tex]\Large \boxed{\sf \Delta = 5}[/tex]

[tex]\Large \boxed{\sf \sf{x = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{-1 \pm \sqrt{5}}{2} \rightarrow \begin{cases}\sf{x_2 = \dfrac{-1 + \sqrt{5}}{2}}\\\\\sf{x_3 = \dfrac{-1 - \sqrt{5}}{2}}\end{cases}}}[/tex]

[tex]\Large \boxed{\boxed{\sf S = \left\{1;\:\dfrac{-1 + \sqrt{5}}{2};\:\dfrac{-1 - \sqrt{5}}{2}\right\}}}[/tex]

[tex]\Large \boxed{\sf x_1 + \dfrac{1}{x_1} = 1 + \dfrac{1}{1} = 2}[/tex]

[tex]\Large \boxed{\sf x_2 + \dfrac{1}{x_2} = \left(\dfrac{-1 + \sqrt{5}}{2}\right) + \left(\dfrac{1}{\dfrac{-1 + \sqrt{5}}{2}}\right) = \sqrt{5}}[/tex]

[tex]\Large \boxed{\sf x_3 + \dfrac{1}{x_3} = \left(\dfrac{-1 - \sqrt{5}}{2}\right) + \left(\dfrac{1}{\dfrac{-1 - \sqrt{5}}{2}}\right) = -\sqrt{5}}[/tex]

[tex]\Large \boxed{\boxed{\sf S = \left\{2;\:\sqrt{5};\:-\sqrt{5}\right\}}}[/tex]