a) \(9=6+3=6+\sqrt{9}\)

\(6+2\sqrt{2}=6+\sqrt{8}\)

\(\sqrt{8}< \sqrt{9}\) nên \(6+\sqrt{8}=6+2\sqrt{2}< 6+\sqrt{9}=9\)

b) \(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}=5+\sqrt{24}\)

\(3^2=9=5+4=5+\sqrt{16}\)

\(\sqrt{16}< \sqrt{24}\Rightarrow3^2< \left(\sqrt{2}+\sqrt{3}\right)^2\Rightarrow3< \sqrt{2}+\sqrt{3}\)

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

\(16=\left(2+2\right)^2=\left(2+\sqrt{4}\right)^2\)

\(\sqrt{4}< \sqrt{5}\Rightarrow2+\sqrt{4}< 2+\sqrt{5}\Rightarrow\left(2+\sqrt{4}\right)^2=16< \left(2+\sqrt{5}\right)^2=9+4\sqrt{5}\)

d) \(\left(\sqrt{11}-\sqrt{3}\right)^2=14-2\sqrt{33}=14-\sqrt{132}\)

\(2^2=14-10=14-\sqrt{100}\)

\(\sqrt{100}< \sqrt{132}\Leftrightarrow-\sqrt{100}>-\sqrt{132}\Leftrightarrow14-\sqrt{100}>14-\sqrt{132}\)

\(\Rightarrow2>\sqrt{11}-\sqrt{3}\)

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. 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

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

2  +  3  và 3

Ta có: 2 + 3 2 = 2 2 . 3 2 =2.3=6

2 2 =4

Vì 6 > 4 nên 2 . 3 2   >   2 2

Suy ra:  2 . 3  > 2 ⇒ 2.  2 . 3  > 2.2 ⇒ 5 + 2.  2 . 3  > 4 + 5

⇒ 5 + 2.  2 . 3  > 9 ⇒ ( √2 + √3)2 > 9

⇒ 2 + 3 2 > 3 2

Vậy  2  +  3  > 3

Giúp mình với 

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

\(2>\sqrt{2}\)

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

2 + 3  và  10

Ta có: 2 + 3 2 = 2 + 2 6 + 3 = 5 + 2 6

10 2  = 10 = 5 + 5

So sánh 26 và 5:

Ta có: 2 6 2 = 2 2 . 6 2  = 4.6 = 24

5 2  = 25

Vì 2 6 2 < 5 2  nên 2 6 < 5

Vậy 5 + 2 6  < 5 + 5 ⇒ 2 + 3 2 < 10 2  ⇒  2 + 3 <  10

11  -  3  và 2

Vì  11  >  3  nên  11  -  3  > 0

Ta có: 11 - 3 2  = 11 - 2 11 . 3  + 3 = 14 - 2 11 . 3

2 2  = 4 = 14 – 10

So sánh 10 và 2 11 . 3  hay so sánh giữa 5 và  11 . 3

Ta có:  5 2  = 25

11 . 3 2 = 11 2 . 3 2  = 11.3 = 33

Vì 25 < 33 nên 5 2 < 11 . 3 2

Suy ra:  5 < 11 . 3 2

Suy ra: 14 – 10 > 14 - 2 11 . 3  ⇒ 11 - 3 2 < 2 2

Vậy  11  -  3  < 2

3 + 2  và  2 + 6

Ta có:  3 + 2 2  = 3 + 4 3 + 4 = 7 + 4 3

2 + 6 2  = 2 + 2 12 + 6 = 8 + 2 4 . 3 ) = 8 + 2. 4 . 3 = 8 + 4 3

Vì 7 + 4 3  < 8 + 4 3  nên  3 + 2 2  <  2 + 6 2

Vậy  3 + 2  <  2 + 6

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 \(\sqrt{2004}\)