Resposta:

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

Explicação passo a passo:

[tex]\mathsf{\dfrac{3}{5}\:.\:\dfrac{(x^2 - 1)}{x^2} = \dfrac{2(x^2 - 1)}{5}}[/tex]

[tex]\mathsf{\dfrac{3}{\not5}\:.\:\dfrac{(x^2 - 1)}{x^2} = \dfrac{2(x^2 - 1)}{\not5}}[/tex]

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

[tex]\mathsf{3x^2 - 3 = 2x^4 - 2x^2}[/tex]

[tex]\mathsf{2x^4 - 5x^2 + 3 = 0}[/tex]

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

[tex]\mathsf{2y^2 - 5y + 3 = 0}[/tex]

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

[tex]\mathsf{\Delta = (-5)^2 - 4.2.3}[/tex]

[tex]\mathsf{\Delta = 25 - 24}[/tex]

[tex]\mathsf{\Delta = 1}[/tex]

[tex]\mathsf{y = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{5 \pm \sqrt{1}}{4} \rightarrow \begin{cases}\mathsf{y' = \dfrac{5 + 1}{4} = \dfrac{6}{4} = \dfrac{3}{2}}\\\\\mathsf{y'' = \dfrac{5 - 1}{4} = \dfrac{4}{4} = 1}\end{cases}}[/tex]

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

[tex]\mathsf{x = \pm\:\sqrt{1}}[/tex]

[tex]\mathsf{x = \pm\:1}[/tex]

[tex]\mathsf{x^2 = \dfrac{3}{2}}[/tex]

[tex]\mathsf{x = \pm\:\sqrt{\dfrac{3}{2}}}[/tex]

[tex]\boxed{\boxed{\mathsf{S = \left\{1;-1;\sqrt{\dfrac{3}{2}};-\sqrt{\dfrac{3}{2}\:}\right\}}}}[/tex]