Question

Any help? Much appreciated ​

Answer 1

Answer:

$\frac{6+\sqrt{27} }{4-\sqrt{3} } = \frac{r+s\sqrt{3} }{13} \\We\ may\ multiply\ the\ numerator\ and\ denominator\ with\ (4+\sqrt{3} ).\\Hence,\\\frac{(6+\sqrt{27})(4+\sqrt{3}) }{13} = \frac{r+s\sqrt{3} }{13} \\Hence,\\24+6\sqrt{3} +4\sqrt{27} +9= r+s\sqrt{3}\\24+6\sqrt{3} +4*3\sqrt{3} } +9= r+s\sqrt{3}\\24+6\sqrt{3} +12\sqrt{3}+9 = r+s\sqrt{3}\\33+18\sqrt{3} = r+s\sqrt{3}\\Hence,\\r=33, s=18$