[tex]\displaystyle (\text x + \frac{1}{\text x})^2-2.(\text x+\frac{1}{\text x}) = \frac{5}{4}[/tex]
Fazendo uma mudança de variável :
[tex]\displaystyle (\text x+\frac{1}{\text x}) = \text y[/tex]
Então temos :
[tex]\displaystyle \text y^2-2\text y = \frac{5}{4}[/tex]
[tex]\displaystyle \text y^2-2\text y +1= \frac{5}{4}+1[/tex]
[tex]\displaystyle (\text y-1)^2 = \frac{9}{4}[/tex]
[tex]\displaystyle \text y-1 = \pm \frac{3}{2}[/tex]
[tex]\displaystyle \text y = 1 + \frac{3}{2} \to \boxed{\text y = \frac{5}{2}}[/tex]
[tex]\displaystyle \text y = 1 - \frac{3}{2} \to \boxed{\text y = \frac{-1}{2}}[/tex]
Desfazendo a troca de variável :
1º [tex]\displaystyle (\text x +\frac{1}{\text x})=\frac{5}{2}[/tex]
multiplicando os dois lados por 2x :
[tex]2.\text x^2 + 2 = 5\text x[/tex]
[tex]2\text x^2 -5\text x + 2 = 0[/tex]
[tex]\displaystyle \text x = \frac{-(-5) \pm\sqrt{(-5)^2-4.2.2}}{2.2}[/tex]
[tex]\displaystyle \text x = \frac{5 \pm\sqrt{25-16}}{4}[/tex]
[tex]\displaystyle \text x = \frac{5 \pm\sqrt{9}}{4}\to \text x = \frac{5 \pm 3}{4}[/tex]
[tex]\displaystyle \text x = \frac{5 +3}{4} \to \text x = \frac{8}{4}[/tex]
[tex]\boxed{\displaystyle \text x = 2}[/tex]
[tex]\displaystyle \text x = \frac{5 -3}{4} \to \text x = \frac{2}{4}[/tex]
[tex]\boxed{\displaystyle \text x = \frac{1}{2}}[/tex]
2º [tex]\displaystyle (\text x + \frac{1}{\text x }) = \frac{-1}{2}[/tex]
multiplicando os dois lados por 2x :
[tex]2\text x^2+2 = -\text x[/tex]
[tex]2\text x^2+\text x +2 = 0[/tex]
[tex]\displaystyle \text x =\frac{-1\pm\sqrt{1^2-4.2.2 }}{2.2}\to \text x = \frac{1\pm\sqrt{-15}}{4}[/tex]
Deu raiz negativa, logo não pertence aos reais. Então não convém.
Portanto :
[tex]\huge\boxed{\ \text S : \text{ x = 2} \ \ ; \ \ \text x = \frac{1}{2}\ }\ }\checkmark[/tex]
