Ta có : \(\sqrt{17}+\sqrt{26}+1

Huỳnh Đức Lê viết sai dấu rồi

Sao mk chỉ thấy 2 câu trả lời thôi vậy

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

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

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

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

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

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

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