What is the fastest method to find which of $\frac {3\sqrt {3}-4}{7-2\sqrt {3}} $ and $\frac {3\sqrt {3}-8}{1-2\sqrt {3}} $ is bigger manually? What is the fastest method to find which number is bigger manually?
$\frac {3\sqrt {3}-4}{7-2\sqrt {3}} $ or $\frac {3\sqrt {3}-8}{1-2\sqrt {3}} $

I would multiply by the conjugate of denominator $$a=\frac{3 \sqrt{3}-4}{7-2 \sqrt{3}}=\frac{1}{37} \left(13 \sqrt{3}-10\right)$$ $$b=\frac{3 \sqrt{3}-8}{1-2 \sqrt{3}}=\frac{1}{11} \left(13 \sqrt{3}-10\right)$$ this makes $a<b$.
