\(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
= \(\sqrt{5+2\sqrt{5}+1}-\sqrt{5-2\sqrt{5}+1}\)
= \(\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}\)
= \(\sqrt{5}+1-\sqrt{5}+1\) ( vì \(\sqrt{5}>1\))
= 2.
Đặt A=\(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
=>\(A=\sqrt{5+2\sqrt{5}+1}-\sqrt{5-2\sqrt{5}+1}\)
\(\Leftrightarrow A=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(\Leftrightarrow A=\left|\sqrt{5}+1\right|-\left|\sqrt{5}-1\right|=\sqrt{5}+1-\left(\sqrt{5}-1\right)\)
\(\Leftrightarrow A=2\)