Respuesta :

[tex]\dfrac{6+ \sqrt{27}}{4-\sqrt3} \\\\\\=\dfrac{\left(6+\sqrt{27}\right)\left(4+\sqrt 3 \right)}{\left(4-\sqrt 3\right) \left(4+\sqrt 3\right)}\\\\\\=\dfrac{24+6\sqrt 3+4\sqrt{27}+\sqrt{27\cdot 3}}{4^2 - \left(\sqrt 3 \right)^2}\\\\\\=\dfrac{24+6\sqrt3+4\sqrt{9\cdot 3} +\sqrt{81}}{16-3}\\\\\\=\dfrac{24+6\sqrt 3 +12\sqrt 3 + 9 }{13}\\\\\\=\dfrac{33+18\sqrt 3}{13}\\\\\\\text{Hence, r = 33 and s = 18}[/tex]