Question

Tính: \(\dfrac{\sqrt{8-4\sqrt{3}}}{\sqrt{\sqrt{6}-\sqrt{2}}}.\sqrt{\sqrt{6}+\sqrt{2}}\)

 

Answer 1

\(=\dfrac{\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}}{\sqrt{\sqrt{6}-\sqrt{2}}}.\sqrt{\sqrt{6}+\sqrt{2}}=\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{\sqrt{6}-\sqrt{2}}}.\sqrt{\sqrt{6}+\sqrt{2}}\)

\(=\sqrt{\sqrt{6}-\sqrt{2}}.\sqrt{\sqrt{6}+\sqrt{2}}=\sqrt{\left(\sqrt{6}-\sqrt{2}\right)\left(\sqrt{6}+\sqrt{2}\right)}\)

\(=\sqrt{4}=2\)

Answer 2

Ta có: \(\dfrac{\sqrt{8-4\sqrt{3}}}{\sqrt{\sqrt{6}-\sqrt{2}}}\cdot\sqrt{\sqrt{6}+\sqrt{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{6}-\sqrt{2}\right)^2}{\sqrt{6}-\sqrt{2}}}\cdot\sqrt{\sqrt{6}+\sqrt{2}}\)

=4