Question

Como resolver essa equação? Alguém pode me ajudar?
A) 3/5(x^2 - 1)/x^2 = 2(x^2-1)/5

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{\dfrac{3}{5}\:.\:\dfrac{(x^2 - 1)}{x^2} = \dfrac{2(x^2 - 1)}{5}}$

$\mathsf{\dfrac{3}{\not5}\:.\:\dfrac{(x^2 - 1)}{x^2} = \dfrac{2(x^2 - 1)}{\not5}}$

$\mathsf{\dfrac{(3x^2 - 3)}{x^2} = 2x^2 - 2}$

$\mathsf{3x^2 - 3 = 2x^4 - 2x^2}$

$\mathsf{2x^4 - 5x^2 + 3 = 0}$

$\mathsf{y = x^2}$

$\mathsf{2y^2 - 5y + 3 = 0}$

$\mathsf{\Delta = b^2 - 4.a.c}$

$\mathsf{\Delta = (-5)^2 - 4.2.3}$

$\mathsf{\Delta = 25 - 24}$

$\mathsf{\Delta = 1}$

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

$\mathsf{x^2 = 1}$

$\mathsf{x = \pm\:\sqrt{1}}$

$\mathsf{x = \pm\:1}$

$\mathsf{x^2 = \dfrac{3}{2}}$

$\mathsf{x = \pm\:\sqrt{\dfrac{3}{2}}}$

$\boxed{\boxed{\mathsf{S = \left\{1;-1;\sqrt{\dfrac{3}{2}};-\sqrt{\dfrac{3}{2}\:}\right\}}}}$