Question

comparati numerele a= radical din 2 si b= 1/radical din 3+radical din 2

Answer 1

$a= \sqrt{2} \\ \\ b=\frac{1}{ \sqrt{3}+ \sqrt{2} } = \frac{\sqrt{3}- \sqrt{2}}{(\sqrt{3}- \sqrt{2})(\sqrt{3}+ \sqrt{2}} )= \frac{\sqrt{3}- \sqrt{2}}{(\sqrt{3}- \sqrt{2})(\sqrt{3}+ \sqrt{2}} )= \sqrt{3}- \sqrt{2} \\ \\ a= \sqrt{2} =1,4141135> b= \sqrt{3} - \sqrt{2}=0,31$

a>b