Chứng minh rằng : [tex]\fn_cm \frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}[/tex] là 1 số nguyên
$\dfrac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}=\dfrac{2\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}+\sqrt{2}}
\\=\dfrac{2\sqrt{3+\sqrt{5-\sqrt{(2\sqrt{3}+1)^2}}}}{\sqrt{6}+\sqrt{2}}=\dfrac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}+\sqrt{2}}
\\=\dfrac{2\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{6}+\sqrt{2}}=\dfrac{2\sqrt{3+\sqrt{(\sqrt{3}-1)^2}}}{\sqrt{6}+\sqrt{2}}=\dfrac{2\sqrt{3+\sqrt{3}-1}}{\sqrt{6}+\sqrt{2}}
\\=\dfrac{2\sqrt{2+\sqrt{3}}}{\sqrt{6}+\sqrt2}=\dfrac{\sqrt2\sqrt{4+2\sqrt{3}}}{\sqrt{6}+\sqrt2}=\dfrac{\sqrt2\sqrt{(\sqrt3+1)^2}}{\sqrt2(\sqrt3+1)}
\\=\dfrac{\sqrt2(\sqrt3+1)^2}{\sqrt2(\sqrt3+1)}=1$
[tex]\fn_cm \sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}[/tex]
$\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{(2+\sqrt{3})^2}}}}
\\=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10(2+\sqrt{3})}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt3}}}
\\=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{(5-\sqrt{3})^2}}}
\\=\sqrt{4+\sqrt{5\sqrt{3}+5(5-\sqrt{3})}}=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt3}}
\\=\sqrt{4+\sqrt{25}}=\sqrt{4+5}=\sqrt{9}=3$
[tex]\fn_cm \left ( \sqrt{3}-1 \right ).\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{}\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}[/tex]
$(\sqrt{3}-1).\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}
\\=(\sqrt{3}-1).\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}}
\\=(\sqrt{3}-1).\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{(4-\sqrt{2})^2}}}}
\\=(\sqrt{3}-1).\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+4-\sqrt{2}}}}
\\=(\sqrt{3}-1).\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}
\\=(\sqrt{3}-1).\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{(\sqrt3+1)^2}}}
\\=(\sqrt{3}-1).\sqrt{6+2.\sqrt{6-2(\sqrt3+1)}}
\\=(\sqrt{3}-1).\sqrt{6+2.\sqrt{4-2\sqrt3}}
\\=(\sqrt{3}-1).\sqrt{6+2.\sqrt{(\sqrt3-1)^2}}
\\=(\sqrt{3}-1).\sqrt{6+2(\sqrt3-1)}
\\=(\sqrt{3}-1).\sqrt{4+2\sqrt3}
\\=(\sqrt{3}-1).\sqrt{(\sqrt3+1)^2}
\\=(\sqrt{3}-1).(\sqrt3+1)
\\=3-1=2$
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