a) \(6=\sqrt[3]{6^3}=\sqrt{216}>\sqrt[3]{208}=2\sqrt[3]{26}\)

b) \(2\sqrt[3]{6}=\sqrt[3]{2^3.6}=\sqrt[3]{48}>\sqrt[3]{47}\)

a) Ta có:

\(2=1+1=1+\sqrt{1}\)

Mà: \(1< 2\Rightarrow\sqrt{1}< \sqrt{2}\)

\(\Rightarrow1+\sqrt{1}< \sqrt{2}+1\)

\(\Rightarrow2< \sqrt{2}+1\)

b) Ta có:

\(1=2-1=\sqrt{4}-1\)

Mà: \(4>3\Rightarrow\sqrt{4}>\sqrt{3}\)

\(\Rightarrow\sqrt{4}-1>\sqrt{3}-1\)

\(\Rightarrow1>\sqrt{3}-1\)

c) Ta có:

\(10=2\cdot5=2\sqrt{25}\)

Mà: \(25< 31\Rightarrow\sqrt{25}< \sqrt{31}\)

\(\Rightarrow2\sqrt{25}< 2\sqrt{31}\)

\(\Rightarrow10< 2\sqrt{31}\)

d) Ta có:

\(-12=-3\cdot4=-3\sqrt{16}\)

Mà: \(16>11\Rightarrow\sqrt{16}>\sqrt{11}\)

\(\Rightarrow-3\sqrt{16}< -3\sqrt{11}\)

\(\Rightarrow-12< -3\sqrt{11}\)

a: \(\left(\sqrt{7}+\sqrt{15}\right)^2=22+2\sqrt{105}=7+15+2\sqrt{105}\)

\(7^2=49=7+42\)

mà \(15+2\sqrt{105}< 42\)

nên \(\sqrt{7}+\sqrt{15}< 7\)

b: \(\left(\sqrt{2}+\sqrt{11}\right)^2=13+2\sqrt{22}\)

\(\left(5+\sqrt{3}\right)^2=28+10\sqrt{3}=13+15+10\sqrt{3}\)

mà \(2\sqrt{22}< 15+10\sqrt{3}\)

nên \(\sqrt{2}+\sqrt{11}< 5+\sqrt{3}\)

a)Ta có:\(\sqrt{17}>\sqrt{16}\)

             \(\sqrt{26}>\sqrt{25}\)

\(\implies\) \(\sqrt{17}+\sqrt{26}>\sqrt{16}+\sqrt{25}\)

\(\implies\) \(\sqrt{17}+\sqrt{26}+1>\sqrt{16}+\sqrt{25}+1=4+5+1=10\)

Mà \(\sqrt{100}=10\) \(\implies\) \(\sqrt{17}+\sqrt{26}+1>\sqrt{100}\)

Mà \(\sqrt{100}>\sqrt{99}\) \(\implies\) \(\sqrt{17}+\sqrt{26}+1>\sqrt{99}\)

b)Ta có:\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+....+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}=100.\frac{1}{\sqrt{100}}\)

\(\implies\) \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+....+\frac{1}{\sqrt{100}}>\frac{1}{10}.100=10\)

\(\implies\) \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+....+\frac{1}{\sqrt{100}}>10\left(đpcm\right)\)

b: Ta có: \(4\sqrt{5}=\sqrt{4^2\cdot5}=\sqrt{80}\)

\(5\sqrt{3}=\sqrt{5^2\cdot3}=\sqrt{75}\)

mà 80>75

nên \(4\sqrt{5}>5\sqrt{3}\)

Ta có:

\(\sqrt{99}< \sqrt{100}=10\)

\(\sqrt{17}+\sqrt{26}+1>\sqrt{16}+\sqrt{25}+1=10\)

Vậy \(\sqrt{17}+\sqrt{26}+1>\sqrt{99}\)

ʇɐɥʇ ɥuɐɹ uɐq ɔɐɔ ɐl ƃunp ıɥʇ ʎɐp uǝp ɔonp ɔop uɐq ɔɐɔ ɐl ʇǝıq ɥuıɯ ƃunɥu 'ɔonp ɔop ıoɯ ıɐl ɔonƃu ʎɐox ıɐɥd ɐʌ ɔop oɥʞ ɐl ʇɐɹ ıɥʇ ʎɐu ǝɥʇ ʇǝıʌ ɐl ʇǝıq ɥuıɯ

ʎɐʌ ƃunp

√17 + √26 + 1 và √99 
Ta có: √17 > √16 (1) 
√26 > √25 (2) 
Từ (1) và (2) => √17 + √26 + 1 > √16 + √25 + 1 
=> √17 + √26 + 1 > 4 + 5 + 1 
=> √17 + √26 + 1 > 10 
=> √17 + √26 + 1 > √100 
Do √100 > √99 
=> √17 + √26 + 1 > √99 
 

Ta có 

\(\sqrt{17}+\sqrt{26}+1>\sqrt{16}+\sqrt{25}+1=4+5+1=10=\sqrt{100}\)(1)

Mà \(\sqrt{99}< \sqrt{100}\)(2)

Từ (1)(2) \(\Rightarrow\sqrt{17}+\sqrt{26}+1>\sqrt{99}\)

P/s tham khảo nha

\(\sqrt{7}+\sqrt{15}<\sqrt{9}+\sqrt{16}=3+4=7\Rightarrow\sqrt{7}+\sqrt{15}<7\)

\(\sqrt{2}+\sqrt{11}<\sqrt{3}+\sqrt{25}=\sqrt{3}+5\Rightarrow\sqrt{2}+\sqrt{11}<\sqrt{3}+5\)

\(\sqrt{17}+\sqrt{26}+1>\sqrt{16}+\sqrt{25}+1=4+5+1=10\Rightarrow\sqrt{17}+\sqrt{26}+1>10\)

\(\sqrt{99}<\sqrt{100}=10\Rightarrow\sqrt{99}<10\)

Nên  \(\sqrt{17}+\sqrt{26}+1>10\)