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Bài 1: Ta có:
\(\frac{\sqrt{8-4\sqrt{3}}}{\sqrt{\sqrt{6}-\sqrt{2}}}\sqrt{\sqrt{6}+\sqrt{2}}=\frac{\sqrt{8-4\sqrt{3}}}{\sqrt{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})}}(\sqrt{6}+\sqrt{2})\)
\(=\frac{\sqrt{8-4\sqrt{3}}}{\sqrt{6-2}}(\sqrt{6}+\sqrt{2})\)
\(=\frac{\sqrt{6+2-2\sqrt{6.2}}}{2}(\sqrt{6}+\sqrt{2})\)
\(=\frac{\sqrt{(\sqrt{6}-\sqrt{2})^2}}{2}(\sqrt{6}+\sqrt{2})\)
\(=\frac{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})}{2}=\frac{6-2}{2}=2\)
Bài 2:
\(A=\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
\(\Rightarrow A^2=8+2\sqrt{10+2\sqrt{5}}+8-2\sqrt{10+2\sqrt{5}}+2\sqrt{(8+2\sqrt{10+2\sqrt{5}})(8-2\sqrt{10+2\sqrt{5}})}\)
\(=16+2\sqrt{8^2-(2\sqrt{10+2\sqrt{5}})^2}\)
\(=16+2\sqrt{64-4(10+2\sqrt{5})}\)
\(=16+2\sqrt{24-8\sqrt{5}}=16+2\sqrt{20+4-2\sqrt{20.4}}\)
\(=16+2\sqrt{(\sqrt{20}-\sqrt{4})^2}\)
\(=16+2(\sqrt{20}-2)=12+2\sqrt{20}=10+2+2\sqrt{10.2}=(\sqrt{10}+\sqrt{2})^2\)
\(\Rightarrow A=\sqrt{10}+\sqrt{2}\)
\(C=\sqrt{\left(8+2\sqrt{10+2\sqrt{5}}\right).\left(8-2\sqrt{10+2\sqrt{5}}\right)}=\sqrt{\left(8^2-\left(2\sqrt{10+2\sqrt{5}}\right)^2\right)=\sqrt{64-4\left(10+2\sqrt{5}\right)}}\)
\(C=\sqrt{64-40-8\sqrt{5}}=\sqrt{24-8\sqrt{5}}\)
\(C=\sqrt{20-2.2.2\sqrt{5}+4}=\sqrt{\left(2\sqrt{5}-2\right)^2}\)
\(C=2\sqrt{5}-2=2\left(\sqrt{5}-1\right)\)
\(\frac{5}{\sqrt{2}-7}-\frac{4}{3\sqrt{2}+5}-\frac{7}{4-5\sqrt{2}}\)
\(A^2=8+2\sqrt{10+2\sqrt{5}}+8-2\sqrt{10+2\sqrt{5}}+2.\sqrt{\left(8+2\sqrt{10+2\sqrt{5}}\right)\left(8-2\sqrt{10+2\sqrt{5}}\right)}\)
\(A^2=16+2.\sqrt{8^2-\left(2\sqrt{10+2\sqrt{5}}\right)^2}=16+2.\sqrt{24-8\sqrt{5}}=16+4.\sqrt{6-2\sqrt{5}}\)
\(A^2=16+4.\sqrt{\left(\sqrt{5}-1\right)^2}=16+4.\left(\sqrt{5}-1\right)=12+4\sqrt{5}\)
=> A = \(\sqrt{12+4\sqrt{5}}=\sqrt{2}\sqrt{6+2\sqrt{5}}=\sqrt{2}.\left(\sqrt{5}+1\right)=\sqrt{10}+\sqrt{2}\)