1. rationalize

$\displaystyle \displaystyle\ A=\frac{6}{\sqrt{3-\sqrt{3}}}=\sqrt{\frac{36}{3-\sqrt{3}}}=\sqrt{\frac{36(3+\sqrt{3})}{(3-\sqrt{3})(3+\sqrt{3})}$
$\displaystyle \displaystyle\sqrt{\frac{36(3+\sqrt{3})}{9+3\sqrt{ 3}-3\sqrt{3}-3}}=\sqrt{\frac{36(3+\sqrt{3})}{9-3}}$
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