Question

1.\(\dfrac{\sqrt{8-4\sqrt{3}}}{\sqrt{\sqrt{6}-\sqrt{2}}}\cdot\sqrt{\sqrt{6}+\sqrt{2}}\)

2.\(\sqrt{16-5\sqrt{7}}\left(5\sqrt{2}+\sqrt{14}\right)+\dfrac{6}{3+\sqrt{10}}\)

Answer 1

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:

\(\frac{\sqrt{8-4\sqrt{3}}}{\sqrt{\sqrt{6}-\sqrt{2}}}.\sqrt{\sqrt{6}+\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{\sqrt{\sqrt{6}-\sqrt{2}}}.\sqrt{\sqrt{6}+\sqrt{2}}=\sqrt{\sqrt{6}-\sqrt{2}}.\sqrt{\sqrt{6}+\sqrt{2}}\)

\(=\sqrt{(\sqrt{6})^2-(\sqrt{2})^2}=\sqrt{6-2}=2\)

Answer 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:

\(\sqrt{16-5\sqrt{7}}(5\sqrt{2}+\sqrt{14})+\frac{6}{3+\sqrt{10}}=\frac{5-\sqrt{7}}{\sqrt{2}}.\sqrt{2}(5+\sqrt{7})+\frac{6(3-\sqrt{10})}{(3+\sqrt{10})(3-\sqrt{10})}\)

\(=(5-\sqrt{7})(5+\sqrt{7})+\frac{18-6\sqrt{10}}{3^2-10}=25-7+(-18+6\sqrt{10})\)

\(=6\sqrt{10}\)