Giải N : (3$\sqrt{2}+\sqrt{6}$ ).$(\sqrt{6-3\sqrt{3}}$ M : $\sqrt{3+\sqrt{13+\sqrt{48}}}$ 12/08/2021 Bởi Arianna Giải N : (3$\sqrt{2}+\sqrt{6}$ ).$(\sqrt{6-3\sqrt{3}}$ M : $\sqrt{3+\sqrt{13+\sqrt{48}}}$
Đáp án: $\rm N=(3\sqrt2+\sqrt6).\sqrt{6-3\sqrt3}\\=\sqrt2(3+\sqrt3).\sqrt{6-3\sqrt3}\\=(3+\sqrt3).\sqrt{12-6\sqrt3}\\=(3+\sqrt3).\sqrt{9-2.3.\sqrt3+3}\\=(3+\sqrt3).\sqrt{(3-\sqrt3)^2}\\=(3+\sqrt3)(3-\sqrt3)\\=9-3\\=6\\Vậy \,\, N=6\\M=\sqrt{3+\sqrt{13+\sqrt48}}\\=\sqrt{3+\sqrt{12+2.2\sqrt3.1+1}}\\=\sqrt{3+\sqrt{(2\sqrt3+1)^2}}\\=\sqrt{3+2\sqrt3+1}\\=\sqrt{4+2\sqrt3}\\=\sqrt{3+2\sqrt3+1}\\=\sqrt{(\sqrt3+1)^2}\\=\sqrt3+1\\Vậy\,\ M=\sqrt3+1$ Bình luận
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Đáp án:
$\rm N=(3\sqrt2+\sqrt6).\sqrt{6-3\sqrt3}\\=\sqrt2(3+\sqrt3).\sqrt{6-3\sqrt3}\\=(3+\sqrt3).\sqrt{12-6\sqrt3}\\=(3+\sqrt3).\sqrt{9-2.3.\sqrt3+3}\\=(3+\sqrt3).\sqrt{(3-\sqrt3)^2}\\=(3+\sqrt3)(3-\sqrt3)\\=9-3\\=6\\Vậy \,\, N=6\\M=\sqrt{3+\sqrt{13+\sqrt48}}\\=\sqrt{3+\sqrt{12+2.2\sqrt3.1+1}}\\=\sqrt{3+\sqrt{(2\sqrt3+1)^2}}\\=\sqrt{3+2\sqrt3+1}\\=\sqrt{4+2\sqrt3}\\=\sqrt{3+2\sqrt3+1}\\=\sqrt{(\sqrt3+1)^2}\\=\sqrt3+1\\Vậy\,\ M=\sqrt3+1$