Ta có: \(A=\frac{\sqrt{3-\sqrt5}\left(3+\sqrt5\right)}{\sqrt{10}+\sqrt2}\)

\(=\frac{\sqrt{6-2\sqrt5}\left(3+\sqrt5\right)}{\sqrt2\cdot\sqrt2\left(\sqrt5+1\right)}=\frac{\left(\sqrt5-1\right)\left(3+\sqrt5\right)}{2\left(\sqrt5+1\right)}\)

\(=\frac{3\sqrt5+5-3-\sqrt5}{2\left(\sqrt5+1\right)}=\frac{2\sqrt5+2}{2\sqrt5+2}=1\)

Ta có: \(3-\sqrt5-1=2-\sqrt5<0\)

=>\(3-\sqrt5<1\)

=>\(\sqrt{3-\sqrt5}<1\)

=>B<1

Do đó: B<A

ta có \(\left(\sqrt{5\sqrt{3}}\right)^4=75\)

          \(\left(\sqrt{3\sqrt{5}}\right)^4=45\)

 \(\Rightarrow\sqrt{5\sqrt{3}}>\sqrt{3\sqrt{5}}\left(75>45\right)\)

Mình thấy cả 2 con trên đều bằng nhau .

\(\sqrt{\sqrt{6+\sqrt{20}}}=\sqrt{\sqrt{5+2\sqrt{5}+1}}=\sqrt{\sqrt{\left(\sqrt{5}+1\right)^2}}=\sqrt{\sqrt{5}+1}< \sqrt{\sqrt{6}+1}\)

\(\sqrt{2}\)+3=3+\(\sqrt{2}\)

\(\sqrt{3}\)+2=2+\(\sqrt{3}\)

\(\Rightarrow\)\(\sqrt{2}\)+3>\(\sqrt{3}\)+2

m kmnhbk5htb ,k55555555555555555555555555555555555e,

\(\sqrt{\sqrt{6+\sqrt{20}}}=\sqrt{\sqrt{6+2\sqrt{5}}}=\sqrt{\sqrt{\left(\sqrt{5}+1\right)^2}}=\sqrt{\sqrt{5}+1}\)

Vì \(\sqrt{\sqrt{5}+1}< \sqrt{\sqrt{6}+1}\Rightarrow\sqrt{\sqrt{6+\sqrt{20}}}< \sqrt{1+\sqrt{6}}\)