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

tao đây

Ghi nhầm 

\(\sqrt{3}+1<\sqrt{4}+1=3\)

Vậy 3 > \(\sqrt{3}+1\)

Lời giải:

\(5\sqrt{2}+4\sqrt{5}-16=(\sqrt{50}-7)+(\sqrt{80}-9)\)

\(=\frac{1}{\sqrt{50}+7}-\frac{1}{\sqrt{80}+9}\)

Dễ thấy \(\sqrt{50}+7< \sqrt{80}+9\Rightarrow \frac{1}{\sqrt{50}+7}>\frac{1}{\sqrt{80}+9}\)

\(\Rightarrow 5\sqrt{2}+4\sqrt{5}-16=\frac{1}{\sqrt{50}+7}-\frac{1}{\sqrt{80}+9}>0\)

\(\Rightarrow 5\sqrt{2}+4\sqrt{5}>16\)

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

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 .

Ta có :

\(\sqrt{50}+\sqrt{5}>\sqrt{49}+\sqrt{4}=7+2=9\)

Vậy \(\sqrt{50}+\sqrt{5}>9\)

dấu :>

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