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

Ta có : \(1=3-2=\sqrt{9}-\sqrt{4}\)

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

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

Ghi nhầm 

\(\sqrt{3}+1

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

Có \(\sqrt{\sqrt{6+\sqrt{20}}}=\sqrt{\sqrt{5+2\sqrt{5}+1}}\)

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

                                          \(=\sqrt{1+\sqrt{5}}< \sqrt{1+\sqrt{6}}\)

Vậy \(\sqrt{\sqrt{6+\sqrt{20}}}< \sqrt{1+\sqrt{6}}\)

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

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