Question

(Radical din 6 - radical din 2) : (radical din 3 - 1)

Answer 1

$\frac{ \sqrt{6} - \sqrt{2} }{ \sqrt{3} - 1} = \frac{ \sqrt{3} \times \sqrt{2} - \sqrt{2} }{ \sqrt{3} - 1} = \frac{ \sqrt{2}( \sqrt{3} - 1)}{ \sqrt{3} - 1} = \sqrt{2}$