[tex]P=(12-6\sqrt{3})\sqrt{\frac{3}{14-8\sqrt{3}}}-3\sqrt{2(1-\sqrt{-2\sqrt{3}+4})+2\sqrt{4+2\sqrt{3}}}\\ P=6(2-\sqrt{3})\frac{\sqrt{3}}{2\sqrt{2}-\sqrt{6}}-3\sqrt{2(1-\sqrt{4-2\sqrt{3}})+2\sqrt{4+2\sqrt{3}}}\\ P=6(2-\sqrt{3})\frac{\sqrt{3}}{\sqrt{2}(2-\sqrt{3})}-3\sqrt{2(1-\sqrt{3}+1)+2(1+\sqrt{3})}\\ P=3\sqrt{6}-3\sqrt{4-2\sqrt{3}+2+2\sqrt{3}}=3\sqrt{6}-3\sqrt{6}=0[/tex]
Vậy P = 0