Ta có $\frac{\sqrt{2}}{\sqrt{2-\sqrt{3}}}$ - $\frac{\sqrt{2}}{\sqrt{2+\sqrt{3}}}$
=$\frac{\sqrt{4}}{\sqrt{4-2\sqrt{3}}}$ - $\frac{\sqrt{4}}{\sqrt{4+2\sqrt{3}}}$
=$\frac{\sqrt{4}}{\sqrt{3-2\sqrt{3}+1}}$ - $\frac{\sqrt{4}}{\sqrt{3+2\sqrt{3}+1}}$
=$\frac{\sqrt{4}}{\sqrt{(\sqrt{3}-1)²}}$ - $\frac{\sqrt{4}}{\sqrt{(\sqrt{3}+1)²}}$
=$\frac{\sqrt{4}}{\sqrt{3}-1}$ - $\frac{\sqrt{4}}{\sqrt{3}+1}$
=$\frac{\sqrt{4}(\sqrt{3}+1)}{3-1}$ - $\frac{\sqrt{4}(\sqrt{3}-1)}{3-1}$
=$\frac{\sqrt{12}+4-\sqrt{12}+4}{2}$
=$\frac{8}{2}$
=4
CHO MK XIN TLHN
