Respuesta :

[tex]\dfrac{4}{\sqrt3-\sqrt2}=\dfrac{4}{\sqrt3-\sqrt2}\cdot\dfrac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}=\dfrac{4(\sqrt3+\sqrt2)}{(\sqrt3)^2-(\sqrt2)^2}\\\\=\dfrac{4\sqrt3+4\sqrt2}{3-2}=4\sqrt3+4\sqrt2[/tex]
ACCESS MORE