Question

Ajudaaa
$\sqrt{3} +1 dividido \sqrt{3} -1+ \sqrt{3} -1 dividido \sqrt{3} +1$

Answer 1

$\frac{\sqrt{3}+1}{\sqrt{3}-1}+\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{(\sqrt{3}+1)*(\sqrt{3}+1)}{(\sqrt{3}-1)*(\sqrt{3}+1)}+\frac{(\sqrt{3}-1)*(\sqrt{3}-1)}{(\sqrt{3}+1)*(\sqrt{3}-1)}$

Denominadores iguais, podemos somar os numeradores:

$\frac{(\sqrt{3}+1)*(\sqrt{3}+1)}{(\sqrt{3}-1)*(\sqrt{3}+1)}+\frac{(\sqrt{3}-1)*(\sqrt{3}-1)}{(\sqrt{3}+1)*(\sqrt{3}-1)}=\frac{(\sqrt{3}+1)^{2}+(\sqrt{3}-1)^{2}}{(\sqrt{3}+1)(\sqrt{3}-1)}$

Resolvendo os 3 produtos notáveis (quadrado da soma, quadrado da diferença e produto da soma pela diferença):

$\frac{(\sqrt{3}+1)*(\sqrt{3}+1)}{(\sqrt{3}-1)*(\sqrt{3}+1)}+\frac{(\sqrt{3}-1)*(\sqrt{3}-1)}{(\sqrt{3}+1)*(\sqrt{3}-1)}=\frac{([\sqrt{3}]^{2}+2*\sqrt{3}*1+1^{2})+([\sqrt{3}]^{2}-2*\sqrt{3}*1+1^{2})}{(\sqrt{3})^{2}-1^{2}}\\\\\frac{(\sqrt{3}+1)*(\sqrt{3}+1)}{(\sqrt{3}-1)*(\sqrt{3}+1)}+\frac{(\sqrt{3}-1)*(\sqrt{3}-1)}{(\sqrt{3}+1)*(\sqrt{3}-1)}=\frac{3+2\sqrt{3}+1+3-2\sqrt{3}+1}{3-1}$

$\frac{(\sqrt{3}+1)*(\sqrt{3}+1)}{(\sqrt{3}-1)*(\sqrt{3}+1)}+\frac{(\sqrt{3}-1)*(\sqrt{3}-1)}{(\sqrt{3}+1)*(\sqrt{3}-1)}=\frac{6+2}{2}\\\\\frac{(\sqrt{3}+1)*(\sqrt{3}+1)}{(\sqrt{3}-1)*(\sqrt{3}+1)}+\frac{(\sqrt{3}-1)*(\sqrt{3}-1)}{(\sqrt{3}+1)*(\sqrt{3}-1)}=\frac{8}{2}\\\\\frac{(\sqrt{3}+1)*(\sqrt{3}+1)}{(\sqrt{3}-1)*(\sqrt{3}+1)}+\frac{(\sqrt{3}-1)*(\sqrt{3}-1)}{(\sqrt{3}+1)*(\sqrt{3}-1)}=4$