9. หาค่า $\sqrt{12-\sqrt{24}+\sqrt{39}-\sqrt{104}} - \sqrt{12+\sqrt{24}+\sqrt{39}+\sqrt{104}}$
ให้ $a = \sqrt{12-\sqrt{24}+\sqrt{39}-\sqrt{104}} - \sqrt{12+\sqrt{24}+\sqrt{39}+\sqrt{104}}$ จะได้
$$a^2 = 24 + 2\sqrt{39} - 2\sqrt{(12+\sqrt{39})^2 - (\sqrt{24}+\sqrt{104})^2}$$
$$a^2 = 24 + 2\sqrt{39} - 2\sqrt{55+2\sqrt{624}}$$
$$a^2 = 24 + 2\sqrt{39} - 2\sqrt{(\sqrt{39}+4)^2}$$
$$a^2 = 24 + 2\sqrt{39} - 2(\sqrt{39}+4)$$
$$a^2 = 24 + 2\sqrt{39} - 2\sqrt{39} - 8$$
$$a^2 = 24 - 8 =16$$
$$a = 4,-4$$
เเต่ $\sqrt{12-\sqrt{24}+\sqrt{39}-\sqrt{104}} < \sqrt{12+\sqrt{24}+\sqrt{39}+\sqrt{104}}$ ทำให้ $a<0$ นั่นคือ $a=-4$
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