Question

radical din 3 minus radical din 2 supra radical din 20

Answer 1

$\dfrac{\sqrt{3} - \sqrt{2} }{\sqrt{20} } = \dfrac{\sqrt{3} - \sqrt{2} }{2\sqrt{5} } = \dfrac{(\sqrt{3} - \sqrt{2} ) \cdot \sqrt{5} }{2\sqrt{5} \cdot\sqrt{5} } = \dfrac{\sqrt{15} - \sqrt{10} }{10}$