Question

Tính: a, \(\sqrt{2006+2\sqrt{2005}}-\sqrt{2006-2\sqrt{2005}}\)

b, \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)

Answer 1

\(\sqrt{2006+2\sqrt{2005}}-\sqrt{2006-2\sqrt{2005}}\)

\(=\sqrt{\left(\sqrt{2005}+1\right)^2}-\sqrt{\left(\sqrt{2005}-1\right)^2}\)

\(=\left(\sqrt{2005}+1\right)-\left(\sqrt{2005}-1\right)\)

= 2

M = \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)

\(\Rightarrow\sqrt{2}M\)\(=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)

\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)

\(=\left(\sqrt{7}-1\right)-\left(\sqrt{7}+1\right)\)

= - 2

\(\Rightarrow M=-\sqrt{2}\)