a. Ta có : \(\sqrt{8}< \sqrt{9}\) ( vì 8< 9)

hay \(2\sqrt{2}< 3\)

\(\Rightarrow\) \(2\sqrt{2}+6< 3+6\)

hay \(2\sqrt{2}+6< 9\)

b. Ta có : \(\sqrt{6}>\sqrt{4}\) (vì 6 > 4 )

hay \(\sqrt{2.3}>2\)

\(\Rightarrow\) 2\(\sqrt{2.3}\) > 4

\(\Rightarrow\) 2 + \(2\sqrt{2.3}\) + 3 > 9

hay \(\left(\sqrt{2}+\sqrt{3}\right)^2\)> 9

\(\Rightarrow\) \(\sqrt{2}+\sqrt{3}>3\)

c. Ta có: \(\sqrt{80}>\sqrt{49}\) (vì 80>49)

hay \(4\sqrt{5}\) > 7

\(\Rightarrow\) 9 + \(4\sqrt{5}\) > 16

d. Ta có : \(2\sqrt{33}>2\sqrt{25}\) (vì 33> 25 ) hay \(2\sqrt{23}>2.5\)

\(\Rightarrow\) - \(2\sqrt{33}\) < - 2.5

\(\Rightarrow\) 11 - \(2\sqrt{11.3}\) +3 < 11- 2.5 +3

hay \(\left(\sqrt{11}-\sqrt{3}\right)^2\) < 4

\(\Rightarrow\) \(\sqrt{11}-\sqrt{3}< 2\)

a,  \(1< 2\Rightarrow\sqrt{1}< \sqrt{2}\Rightarrow1+1< \sqrt{2}+1\Rightarrow2< \sqrt{2}+1\)

c, \(4>3=>\sqrt{4}>\sqrt{3}=>\sqrt{4}-1>\sqrt{3}-1\Rightarrow1>\sqrt{3}-1\)

d, \(16>11=>\sqrt{16}>\sqrt{11}\Rightarrow4>\sqrt{11}=>4.\left(-3\right)< \sqrt{11}.\left(-3\right)\)

\(=>-12< -3.\sqrt{11}\) 

a) \(2\sqrt[3]{3}=\sqrt[3]{2^3}.\sqrt[3]{3}=\sqrt[3]{2^3.3}=\sqrt[3]{24}\)

Ta có : \(24>23\), nên \(\sqrt[3]{24}>\sqrt[3]{23}\)

Vậy \(2\sqrt[3]{3}>\sqrt[3]{23}\)

b) Ta có :

\(11=\sqrt[3]{11^3}=\sqrt[3]{1331}\)

Từ đó suy ra \(33< 3\sqrt[3]{1333}\)

Đặt A = \(\sqrt{ }\)2003 + \(\sqrt{ }\)2005 ; B = 2\(\sqrt{ }\)2004
A² = 2003 + 2005 + 2\(\sqrt{ }\)(2003.2005)
= 4008 + 2\(\sqrt{ }\)[(2004-1)(2004+1)]
= 4008 + 2\(\sqrt{ }\)(2004² - 1) < 2.2004 + 2\(\sqrt{ }\)(2004²) = 4.2004 = B²
\(\Rightarrow\) A < B

Ta có: \(2\sqrt{2003.2005}=2\sqrt{2004^2-1}< 2\sqrt{2004^2}\)

\(\Rightarrow\) 2003 + \(2\sqrt{2003.2005}+2005\) < 2003 + 4008 + 2005

hay \(\left(\sqrt{2003}+\sqrt{2005}\right)^2< 8016\)

\(\Rightarrow\) \(\sqrt{2003}+\sqrt{2005}\) < 2 \(\sqrt{2004}\)

Áp dụng bđt \(\frac{\sqrt{a}+\sqrt{b}}{2}< \sqrt{\frac{a+b}{2}}\) (bạn tự c/m) với a = 2003 , b = 2005

được : \(\frac{\sqrt{2003}+\sqrt{2005}}{2}< \sqrt{\frac{2003+2005}{2}}\)

\(\Rightarrow\sqrt{2003}+\sqrt{2005}< 2\sqrt{2004}\)

a, 2020 lớn hơn

a)\(\left(\sqrt{2019.2021}\right)^2=2019.2021=\left(2020-1\right)\left(2020+1\right)=2020^2-1< 2020^2\)

=> \(\sqrt{2019.2021}< 2020\)

b) \(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}>5+2\sqrt{4}=5+2.2=9\)

=> \(\sqrt{2}+\sqrt{3}>3\)

c) \(9+4\sqrt{5}=4+4\sqrt{5}+5=\left(2+\sqrt{5}\right)^2>\left(2+\sqrt{4}\right)^2=\left(2+2\right)^2=16\)

=> \(9+4\sqrt{5}>16\)

d) \(\sqrt{11}-\sqrt{3}>\sqrt{9}-\sqrt{1}=3-1=2\)

=> \(\sqrt{11}-\sqrt{3}>2\)

Giúp mình với 

\(\sqrt{5-3}=\sqrt{2}\)

\(2>\sqrt{2}\)

\(\Leftrightarrow2>\sqrt{5-3}\)