\(a.\) Xét : \(\sqrt{27}-\sqrt{2}-\sqrt{3}=3\sqrt{3}-\sqrt{2}-\sqrt{3}=2\sqrt{3}-\sqrt{2}=\sqrt{2}\left(\sqrt{6}-1\right)>0\)
⇒ \(\sqrt{27}-\sqrt{2}>\sqrt{3}\)
\(b.\) Gỉa sử : \(\sqrt{5\sqrt{6}}>\sqrt{6\sqrt{5}}\)
⇔ \(5\sqrt{6}>6\sqrt{5}\) ⇔ \(\sqrt{30}\left(\sqrt{5}-\sqrt{6}\right)< 0\)
⇒ \(\sqrt{5\sqrt{6}}< \sqrt{6\sqrt{5}}\)