x \(\ge\) 4 ; x \(\le\) -2

đề đọc không hiểu

b) \(\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}\) + \(\dfrac{2-\sqrt{3}}{\sqrt{2}+\sqrt{2-\sqrt{3}}}\)

= \(\dfrac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{4+2\sqrt{3}}}\) + \(\dfrac{2\sqrt{2}-\sqrt{6}}{2+\sqrt{4-2\sqrt{3}}}\)

= \(\dfrac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}\) + \(\dfrac{2\sqrt{2}-\sqrt{6}}{2+\sqrt{\left(\sqrt{3}-1\right)^2}}\)

= \(\dfrac{2\sqrt{2}+\sqrt{6}}{3+\sqrt{3}}\) + \(\dfrac{2\sqrt{2}-\sqrt{6}}{\sqrt{3}+1}\)

= \(\dfrac{2\sqrt{2}+\sqrt{6}}{\sqrt{3}\left(\sqrt{3}+1\right)}\) + \(\dfrac{2\sqrt{2}-\sqrt{6}}{\sqrt{3}+1}\)

= \(\dfrac{2\sqrt{2}+\sqrt{6}+\left(2\sqrt{2}-\sqrt{6}\right)\sqrt{3}}{\sqrt{3}\left(\sqrt{3}+1\right)}\)

= \(\dfrac{2\sqrt{2}+\sqrt{6}+2\sqrt{6}-3\sqrt{2}}{\sqrt{3}\left(\sqrt{3}+1\right)}\)

= \(\dfrac{3\sqrt{6}-\sqrt{2}}{\sqrt{3}\left(\sqrt{3}+1\right)}\)

câu b giải sao vậy ?