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

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

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

\(\sqrt{8}\)-\(\sqrt{5}\)<1

Ta có : \(1=3-2=\sqrt{9}-\sqrt{4}\)

Vì \(\left\{{}\begin{matrix}\sqrt{9}>\sqrt{8}\\\sqrt{4}< \sqrt{5}\end{matrix}\right.\Rightarrow}\left\{{}\sqrt{8}-\sqrt{5}< \sqrt{9}-\sqrt{4}=1}\)

Ghi nhầm 

\(\sqrt{3}+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 .

\(\sqrt{5\sqrt{3}}\)và \(\sqrt{3\sqrt{5}}\)

\(\sqrt{5\sqrt{3}}\)= 2,94283................

 \(\sqrt{3\sqrt{5}}\)= 2,59002...............

Ta thấy 2,94283......... > 2,59002.........Suy ra \(\sqrt{5\sqrt{3}}\) > \(\sqrt{3\sqrt{5}}\)

1

\(5\) và   \(\sqrt{2}\)

\(5>\sqrt{2}\)

2

\(6\) và   \(\sqrt{81}\)

\(6< \sqrt{81}\)

3

\(\sqrt{81}\) và    \(9\)

\(\sqrt{81}=9\)

5 > \(\sqrt{2}\)

6 < \(\sqrt{81}\)

\(\sqrt{81}\)= 9

Nếu mình đúng thì các bạn k mình nhé

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

Cách 1: Theo casio ta có:

+ \(\sqrt{3}+\sqrt{7}\approx4,378\)

+ \(\sqrt{19}\approx4,36\)

=> \(\sqrt{3}+\sqrt{7}>\sqrt{19}\)

Cách 2: Ta có: \(\left(\sqrt{3}+\sqrt{7}\right)^2=3+7+2.\sqrt{21}=10+\sqrt{84}\)

\(\left(\sqrt{19}\right)^2=19=10+\sqrt{81}\)

Vì \(10+\sqrt{84}>10+\sqrt{81}\)

=> \(\left(\sqrt{3}+\sqrt{7}\right)^2>\left(\sqrt{19}\right)^2\)

=> \(\sqrt{3}+\sqrt{7}>\sqrt{19}\)

Ta có: \(\left(\sqrt{3}+\sqrt{7}\right)^2=10+2\sqrt{21}>10+2\sqrt{20,25}=10+2\sqrt{\left(4,5\right)^2}=10+2.4,5=10+9=19=\left(\sqrt{19}\right)^2\)

(Vì 21 > 20,25 > 0 => \(\sqrt{21}>\sqrt{20,25}\))

Mà 2 biểu thức so sánh đều dương

=>\(\sqrt{3}+\sqrt{7}>\sqrt{19}\).