Đặt \(a=\sqrt{2006}-\sqrt{2005};b=\sqrt{2006}+\sqrt{2005}\)
Ta có
\(a=\sqrt{2006}-\sqrt{2005}=\dfrac{\left(\sqrt{2006}-\sqrt{2005}\right)\left(\sqrt{2006}+\sqrt{2005}\right)}{\sqrt{2006}+\sqrt{2005}}=\dfrac{1}{b}\)
\(\RightarrowĐfcm\)
\(\sqrt{2006}-\sqrt{2005}=\dfrac{1}{\sqrt{2006}+\sqrt{2005}}\)
Do đó: \(\sqrt{2006}-\sqrt{2005};\sqrt{2006}+\sqrt{2005}\) là hai số nghịch đảo