Bài 2 tính thì đúng hơn đấy
2.
P = [tex]\frac{2\sqrt{3 + \sqrt{5 - \sqrt{13 + \sqrt{48}}}}}{\sqrt{6} + \sqrt{2}}[/tex]
= [tex]\frac{2\sqrt{3 \sqrt{5 - \sqrt{13 + 2\sqrt{12}}}}}{\sqrt{6} + \sqrt{2}}[/tex]
= [tex]\frac{2\sqrt{3 + \sqrt{5 - \sqrt{12 + 2\sqrt{12} + 1}}}}{\sqrt{6} + \sqrt{2}}[/tex]
= [tex]\frac{2\sqrt{3 + \sqrt{5 - \sqrt{(\sqrt{12} + 1)^2}}}}{\sqrt{6} + \sqrt{2}}[/tex]
= [tex]\frac{2\sqrt{3 + \sqrt{5 - \sqrt{12} - 1}}}{\sqrt{6} + \sqrt{2}}[/tex]
= [tex]\frac{2\sqrt{3 + \sqrt{4 - 2\sqrt{3}}}}{\sqrt{6} + \sqrt{2}}[/tex]
= [tex]\frac{2\sqrt{3 + \sqrt{(\sqrt{3} - 1)^2}}}{\sqrt{6} + \sqrt{2}}[/tex]
= [tex]\frac{2\sqrt{3 + \sqrt{3} - 1}}{\sqrt{6} + \sqrt{2}}[/tex]
= [tex]\frac{2\sqrt{2 + \sqrt{3}}}{\sqrt{6} + \sqrt{2}}[/tex]
= [tex]\frac{\sqrt{2}.\sqrt{4 + 2\sqrt{3}}}{\sqrt{6} + \sqrt{2}}[/tex]
= [tex]\frac{\sqrt{2}.\sqrt{(\sqrt{3} + 1)^2}}{\sqrt{6} + \sqrt{2}}[/tex]
= [tex]\frac{\sqrt{2}(\sqrt{3} + 1)}{\sqrt{6} + \sqrt{2}}[/tex]
= [tex]\frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} + \sqrt{2}}[/tex]
= 1