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a)\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{1}=1}\)
b) \(B=\sqrt{\sqrt{3}-\sqrt{1+\sqrt{21-6\sqrt{12}}}=\sqrt{\sqrt{3}-\sqrt{1+\sqrt{\left(3-2\sqrt{3}\right)^2}}}}=\sqrt{\sqrt{3}-\sqrt{2\sqrt{3}-2}}\)c)
\(C=\sqrt{7+3\sqrt{5}}+\sqrt{3-\sqrt{5}}=\frac{\sqrt{14+6\sqrt{5}}+\sqrt{6-2\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}=\frac{2+2\sqrt{5}}{\sqrt{2}}=\sqrt{2}+\sqrt{10}=\sqrt{2}\left(\sqrt{5}+1\right)\)
\(A=\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}=\sqrt{1}=1\)
\(A=\sqrt[3]{8-\sqrt{60}}+\sqrt[3]{8+\sqrt{60}}\) xem lại đề con này
\(A=\frac{2\sqrt{3+\sqrt{5-\left(2\sqrt{3}+1\right)}}}{\sqrt{6}+\sqrt{2}}=\frac{2\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{6}+\sqrt{2}}=\frac{2\sqrt{3+\sqrt{3}-1}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{4+2\sqrt{3}}}{2\left(\sqrt{3}+1\right)}=\frac{\sqrt{3}+1}{\sqrt{3}+1}=1\)
ta có: A3=\(6\sqrt{3}+10-6\sqrt{3}+10-3\sqrt[3]{\left(6\sqrt{3}+10\right)\left(6\sqrt{3}-10\right)}.\left(\sqrt[3]{6\sqrt{3}+10}-\sqrt[3]{6\sqrt{3}-10}\right)\)
=\(20-3.\sqrt[3]{8}.A\)=\(20-6A\)
do đó A3=20-6A↔A3+6A-20=0↔(A2+2A+10)(A-2)=0
dễ thấy A2+2A+10>0→A=2
b) giống a)
c)giống b)
a) \(\sqrt{6-4\sqrt{2}}+\sqrt{19-6\sqrt{2}}\)\(=\sqrt{4-4\sqrt{2}+2}+\sqrt{18-2.3\sqrt{2}.1+1}=\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-1\right)^2}\)\(=\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-1\right)^2}\)
= / 2 - \(\sqrt{2}\) / + / 3\(\sqrt{2}\) - 1/
= 2 - \(\sqrt{2}\) + 3\(\sqrt{2}\) - 1
= 2\(\sqrt{2}\) + 1
b) \(\sqrt{46-6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(=\sqrt{45-2.3.\sqrt{5}+1}-\sqrt{20-2.3.2.\sqrt{5}+9}\)
\(=\sqrt{\left(3\sqrt{5}-1\right)^2}-\sqrt{\left(2\sqrt{5}-3\right)^2}\)
= / 3\(\sqrt{5}\) - 1/ - / 2\(\sqrt{5}\) - 3/
= 3\(\sqrt{5}\) - 1 - 2\(\sqrt{5}\) + 3
= \(\sqrt{5}\) + 2
c) \(\sqrt{7-2\sqrt{10}}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5-2\sqrt{5}.\sqrt{2}+2}-\sqrt{5-2\sqrt{5}+1}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}\)
= / \(\sqrt{5}\) - \(\sqrt{2}\) / - / \(\sqrt{5}\) - 1 /
= 1 - \(\sqrt{2}\)
a) \(\sqrt{6-4\sqrt{2}}+\sqrt{19-6\sqrt{2}}\)
\(=\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-1\right)^2}\)
\(=2-\sqrt{2}+3\sqrt{2}-1\)
\(=2\sqrt{2}+1\)
\(T=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{9-12\sqrt{5}+20}}}\)
\(T=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(3-2\sqrt{5}\right)^2}}}=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)
\(T=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)