Ta có: \(12>9\)

\(6\sqrt{3}>4\sqrt{5}\)

Do đó: \(12+6\sqrt{3}>9+4\sqrt{5}\)

\(\Leftrightarrow\sqrt{12+6\sqrt{3}}>\sqrt{9+4\sqrt{5}}\)

8*9 lớn hơn 9*8

tk mình nha

8.9=9.8

easy

ngu ak thị thủy

9 + 4 5  và 16

So sánh 4 5  và 5

Ta có: 16 > 5 ⇒  16  >  5  ⇒ 4 >  5

Vì  5  > 0 nên 4.  5  >  5 . 5  ⇒ 4 5  > 5 ⇒ 9 + 4 5  > 5 + 9

Vậy 9 + 4 5  > 16

Ta có:\(4\sqrt{5}-\sqrt{26}=\sqrt{16}.\sqrt{5}-\sqrt{26}\)

\(=\sqrt{80}-\sqrt{26}\)

\(< \sqrt{81}-\sqrt{26}< \sqrt{81}-\sqrt{25}\)

\(=9-5=4\)

Vậy \(4>4\sqrt{5}-\sqrt{26}\)

Ta có : \(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2006}\)

                \(=\frac{2007-1}{2007}+\frac{2008-1}{2008}+\frac{2009-1}{2009}+\frac{2006+3}{2006}\)

                  \(=1-\frac{1}{2007}+1-\frac{1}{2008}+1-\frac{1}{2009}+1+\frac{3}{2006}\)

                  \(=\left(1+1+1+1\right)-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}-\frac{3}{2006}\right)\)

                  \(< 4-\left(\frac{1}{2009}+\frac{1}{2009}+\frac{1}{2009}-\frac{3}{2009}\right)\)     

                    \(=4\)

=> A < 4 

Vậy A < 4 

A = \(\dfrac{5^{2020}+1}{5^{2021}+1}\) ⇒ A \(\times\) 10 = 2 \(\times\)5 \(\times\) \(\dfrac{5^{2020}+1}{5^{2021}+1}\) =2\(\times\) \(\dfrac{5^{2021}+5}{5^{2021}+1}\)

10A =2 \(\times\) \(\dfrac{5^{2021}+5}{5^{2021}+1}\) = 2 \(\times\)(1 + \(\dfrac{4}{5^{2021}+1}\) )= 2 + \(\dfrac{8}{5^{2021}+1}\) >2

B = \(\dfrac{10^{2019}+1}{10^{2020}+1}\) ⇒ B \(\times\) 10 = 10 \(\times\) \(\dfrac{10^{2019}+1}{10^{2020}+1}\)= \(\dfrac{10^{2020}+10}{10^{2020}+1}\)

10B = \(\dfrac{10^{2020}+10}{10^{2020}+1}\) = 1 + \(\dfrac{9}{10^{2020}+1}\) < 2

10A > 2 > 10B ⇒ 10A>10B ⇒ A>B

\(x^2=3+5+2\sqrt{15}=8+\sqrt{60}\)

\(y^2=2+6+2\sqrt{12}=8+\sqrt{48}\)

Mà \(60>48\Rightarrow\sqrt{60}>\sqrt{48}\Rightarrow8+\sqrt{10}>8+\sqrt{48}\)

\(\Rightarrow x^2>y^2\Rightarrow x>y\) (do x;y đều dương)

A. ta  có \(5=\sqrt{25}\)

vì \(\sqrt{25}< \sqrt{29}\)

suy ra \(5< \sqrt{29}\)

k cho mk nha

waaaaaaaakholuon

5=\(\sqrt{25}\)<\(\sqrt{29}\)(do 25<29)

\(3\sqrt{2}\)=\(\sqrt{18}\);\(2\sqrt{3}\)=\(\sqrt{12}\)

Nhu vay \(3\sqrt{2}\)>\(2\sqrt{3}\)