Question

Como resolvo: $\frac{x+ \sqrt{3} }{\sqrt{x}+ \sqrt{x+ \sqrt{x} 3} } + \frac{x- \sqrt{3} }{ \sqrt{x}- \sqrt{x- \sqrt{3} } } = \sqrt{x}$

Answer 1

$\frac{x+ \sqrt{3} }{x+ \sqrt{x+ \sqrt{3} } } + \frac{x - \sqrt{3} }{ \sqrt{x} - \sqrt{x- \sqrt{3} } } \ = \ \sqrt{x}$

$Devemos \ considerar : \left\{\begin{matrix} x\geq 0 \\ x + \sqrt{3} \geq 0 \\ x - \sqrt{3} \geq 0 \end{matrix}\right. \ \ \ \Rightarrow \ x \geq + \sqrt{3}$

∴ Vou racionalizar :

$\frac{(x+\sqrt{3})(\sqrt{x}-\sqrt{x+\sqrt{3}})}{ (\sqrt{x}+\sqrt{x+\sqrt{3}})(\sqrt{x}-\sqrt{x+\sqrt{3}}) } + \frac{(x-\sqrt{3})(\sqrt{x}+\sqrt{x-\sqrt{3}})}{(\sqrt{x}-\sqrt{x-\sqrt{3}})(\sqrt{x}+\sqrt{x-\sqrt{3}})} = \sqrt{x}$


$\frac{(x+ \sqrt{3})( \sqrt{x} )-( \sqrt{x}+ \sqrt{3} )( \sqrt{x+ \sqrt{3} })}{-\sqrt{3}}+ \frac{(x- \sqrt{3})(\sqrt{x})+(x- \sqrt{3})(\sqrt{x-\sqrt{3}} }{ \sqrt{3} } = \sqrt{x}$

$\frac{-x\sqrt{x}-\sqrt{3x}}{\sqrt{3}} + \frac{\sqrt{(x+\sqrt{3})^3}}{\sqrt{3}} + \frac{x\sqrt{x}-\sqrt{3x}}{\sqrt{3}}+ \frac{\sqrt{(x+\sqrt{3})^3}}{\sqrt{3}} = \sqrt{x}$

$\frac{-2\sqrt{3x}}{\sqrt{3}} + \frac{ \sqrt{(x+\sqrt{3})^3} }{\sqrt{3}} + \frac{ \sqrt{(x-\sqrt{3})^3} }{\sqrt{3}} = \sqrt{x}$

$\sqrt{(x+\sqrt{3})^3} + \sqrt{(x-\sqrt{3})^3} = 3\sqrt{3x}$
$(x+\sqrt{3})^3+(x-\sqrt{3})^3+2\sqrt{[(x+\sqrt{3}).(x-\sqrt{3})]^3}= 27x$
$2\sqrt{(x^2-3)^3}= -2x^3+9x$
$4.(x^2-3)^3 = 4x^6-36x^4+81x^2$
$27x^2 = 108$
$x^2 = 4$

$x^2 = 4 \Rightarrow \left\{\begin{matrix} x = -2 & \ \ (rejeitado) \\ ou & \\ x = 2 & \end{matrix}\right.$

$O \ conjunto \ solu\c{c}\tilde{a}o \ \acute{e} \ : \ S = \left \{ 2 \right \}$