Câu 1:
Có: \(8-4\sqrt{3}=8-2\sqrt{12}=6+2-2\sqrt{6.2}=(\sqrt{6}-\sqrt{2})^2\)
\(\Rightarrow \sqrt{8-4\sqrt{3}}=\sqrt{6}-\sqrt{2}\)
Do đó:
\(\frac{\sqrt{8-4\sqrt{3}}}{\sqrt{\sqrt{6}-\sqrt{2}}}.\sqrt{\sqrt{6}+\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{\sqrt{\sqrt{6}-\sqrt{2}}}.\sqrt{\sqrt{6}+\sqrt{2}}=\sqrt{\sqrt{6}-\sqrt{2}}.\sqrt{\sqrt{6}+\sqrt{2}}\)
\(=\sqrt{(\sqrt{6})^2-(\sqrt{2})^2}=\sqrt{6-2}=2\)
Câu 2:
\(16-5\sqrt{7}=\frac{32-10\sqrt{7}}{2}=\frac{32-2\sqrt{175}}{2}=\frac{25+7-2\sqrt{25.7}}{2}=\frac{(5-\sqrt{7})^2}{2}\)
\(\Rightarrow \sqrt{16-5\sqrt{7}}=\frac{5-\sqrt{7}}{\sqrt{2}}\)
Do đó:
\(\sqrt{16-5\sqrt{7}}(5\sqrt{2}+\sqrt{14})+\frac{6}{3+\sqrt{10}}=\frac{5-\sqrt{7}}{\sqrt{2}}.\sqrt{2}(5+\sqrt{7})+\frac{6(3-\sqrt{10})}{(3+\sqrt{10})(3-\sqrt{10})}\)
\(=(5-\sqrt{7})(5+\sqrt{7})+\frac{18-6\sqrt{10}}{3^2-10}=25-7+(-18+6\sqrt{10})\)
\(=6\sqrt{10}\)
