\(\dfrac{\left(2+\sqrt{3}\right)\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}\\ =\dfrac{\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}\\ = \sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\\ =\sqrt{2^2-\left(\sqrt{3}\right)^2}=\sqrt{4-3}\\ =1\)