How do I divide a square root of a fraction by another square root of another fraction?

Question

My teacher gave this challenger in class today and I can't figure how to solve it: $\frac{\sqrt{\frac23} - \sqrt{\frac32}}{\sqrt{\frac13} - \sqrt{\frac12}}$

Answer 1

Given,

$\frac{\sqrt{\frac23} - \sqrt{\frac32}}{\sqrt{\frac13} - \sqrt{\frac12}}$

On rationalizing,

$\frac{\sqrt{\frac23} - \sqrt{\frac32}}{\sqrt{\frac13} - \sqrt{\frac12}} \cdot \frac{\sqrt{\frac13} + \sqrt{\frac12}}{\sqrt{\frac13} + \sqrt{\frac12}}$

= $\frac{\sqrt{\frac29} + \sqrt{\frac13} - \sqrt{\frac12} - \sqrt{\frac34}}{\frac{-1}{6}}$

= $-6 \left( \frac{\sqrt2}{3} + \frac{1}{\sqrt 3} - \frac{1}{\sqrt 2} - \frac{\sqrt3}{2} \right)$

= $-6 \left( \frac{2\sqrt2 + 2\sqrt3 - 3\sqrt2 - 3\sqrt3}{6} \right)$

= $- \left( - \sqrt2 - \sqrt3 \right)$

= $\sqrt2 + \sqrt3$

Answer 2

mutiplying numerator and denominator by $\sqrt{6}$ gives: $$ \frac{\sqrt{\frac23} - \sqrt{\frac32}}{\sqrt{\frac13} - \sqrt{\frac12}} = \frac{2-3}{\sqrt{2}-\sqrt{3}} = \frac1{\sqrt{3}-\sqrt{2}}=\sqrt{3}+\sqrt{2} $$