Ta có:

\(\sqrt{48}< \sqrt{49}=7\left(1\right)\)

\(13-\sqrt{35}>13-\sqrt{36}=13-6=7\left(2\right)\)

Từ (1) và (2) :

Suy ra: \(\sqrt{48}< 13-\sqrt{35}\)

\(\)

Ta có:

\(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)

\(=\sqrt{\left(2\sqrt{5}\right)^2+2.2\sqrt{5}.2+4}+\sqrt{5-2.2.\sqrt{5}+4}\)

\(=\sqrt{\left(2\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\)

\(=|2\sqrt{5}+2|+|\sqrt{5}-2|\)

\(=2\sqrt{5}+2+\sqrt{5}-2\)

\(=3\sqrt{5}\)

Vì a >b > 0 nên:

\(\sqrt{b}-\sqrt{a}< 0\)

\(< =>2\sqrt{b}\left(\sqrt{b}-\sqrt{a}\right)< 0\)

\(< =>2\left(b-\sqrt{ab}\right)< 0\)

\(< =>2b-2\sqrt{ab}< 0\)

\(< =>a-2\sqrt{ab}+b-a+b< 0\)

\(< =>a-2\sqrt{ab}+b< a-b\)

\(< =>\left(\sqrt{a}-\sqrt{b}\right)^2< a-b\)

\(=>\sqrt{a}-\sqrt{b}< \sqrt{a-b}\)