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

6+2 2  và 9

Ta có: 9 = 6 + 3

So sánh: 2 2 và 3 vì 2 2  > 0 và 3 > 0

Ta có: 2 2 2 = 2 2 . 2 2 =4.2=8

3 2 = 9

Vì 8 < 9 nên :  2 2 2   <   3 2

Vậy 6+2 2  < 9.

\(\sqrt{3}>\sqrt{2}\Rightarrow2\sqrt{2}\cdot\sqrt{3}>4\Rightarrow3+2\sqrt{3}\sqrt{2}+2>9\)

\(\Rightarrow\left(\sqrt{3}+\sqrt{2}\right)^2>3^2\Rightarrow\sqrt{3}+\sqrt{2}>3\Rightarrow\sqrt{3}>3-\sqrt{2}\)

\(\Rightarrow7+\sqrt{3}>10-\sqrt{2}\)

\(a,\left(\sqrt{\sqrt{3}}\right)^4=3< 4=\left(\sqrt{2}\right)^4\Rightarrow\sqrt{\sqrt{3}}< \sqrt{2}\\ b,\left(\sqrt{2\sqrt{3}}\right)^4=12< 18=\left(\sqrt{3\sqrt{2}}\right)^4\Rightarrow\sqrt{2\sqrt{3}}=\sqrt{3\sqrt{2}}\\ c,\left(2+\sqrt{6}\right)^2=8+4\sqrt{6};5^2=25=8+17;\left(4\sqrt{6}\right)^2=96< 289=17^2\\ \Rightarrow4\sqrt{6}< 17\Rightarrow2+\sqrt{6}< 5\\ d,\left(7-2\sqrt{2}\right)^2=57-28\sqrt{2};4^2=16=57-41;\left(28\sqrt{2}\right)^2=1568< 41^2=1681\\ \Rightarrow28\sqrt{2}< 41\Rightarrow7-2\sqrt{2}>4\\ e,\left(\sqrt{15}+\sqrt{8}\right)^2=23+4\sqrt{30};7^2=49=23+26;\left(4\sqrt{30}\right)^2=240< 676=26^2\\ \Rightarrow4\sqrt{30}< 26\Rightarrow\sqrt{15}+\sqrt{8}< 7\)

\(f,\left(\sqrt{37}-\sqrt{14}\right)^2=51-2\sqrt{518};\left(6-\sqrt{15}\right)^2=51-12\sqrt{15};\left(2\sqrt{518}\right)^2=2072;\left(12\sqrt{15}\right)^2=2160\\ \Rightarrow2\sqrt{518}< 12\sqrt{15}\Rightarrow\sqrt{37}-\sqrt{14}>6-\sqrt{15}\)

Bài 6: 

a: \(15=\sqrt{225}>\sqrt{200}\)

b: \(27=9\sqrt{9}>9\sqrt{5}\)

c: \(-24=-\sqrt{576}< -\sqrt{540}=-6\sqrt{15}\)

Áp dụng bđt bunhia copski ta có:

`(sqrt2+sqrt3)^2<=(1+1)(2+3)`

`<=>(sqrt2+sqrt3)^2<=2.5=10`

`=>sqrt2+sqrt3<=sqrt{10}`

Dấu "=" không xảy ra 

`=>sqrt2+sqrt3<sqrt{10}`

Ta có \(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6};\left(\sqrt{10}\right)^2=10=5+5\)

Mà \(\left(2\sqrt{6}\right)^2=24;5^2=25\)

\(\Rightarrow2\sqrt{6}< 5\Rightarrow\left(\sqrt{2}+\sqrt{3}\right)^2< \left(\sqrt{10}\right)^2\)

\(\Rightarrow\sqrt{2}+\sqrt{3}< \sqrt{10}\)

Ta có \(\left(\sqrt{2018}+\sqrt{2020}\right)^2=4038+2\sqrt{4076360}\) và \(\left(2\sqrt{2019}\right)^2=8076=4038+4038\)

Mà \(\left(2\sqrt{4076360}\right)^2=16305440\) và \(4038^2=16305444\)

\(\Rightarrow2\sqrt{4076360}< 4038\)

\(\Rightarrow\sqrt{2018}+\sqrt{2020}< 2\sqrt{2019}\)

\(\left(\sqrt{2018}+\sqrt{2020}\right)^2=4038+2\cdot\sqrt{2018\cdot2020}\)

\(\left(2\sqrt{2019}\right)^2=8076=4038+4038\)

mà \(2\cdot\sqrt{2018\cdot2020}< 4038\)

nên \(\sqrt{2018}+\sqrt{2020}< 2\sqrt{2019}\)

Ta có: 1 < 2 ⇒  1  <  2  ⇒ 1 <  2

Suy ra: 1 + 1 <  2  + 1

Vậy 2 <  2  + 1

a=1+1/1.2+1+1/2.3+....+1+1/9.10

a=1+1+...+1(9 chữ số 1) +1/1-1/2+1/2-1/3+..+1/9-1/10

a=9+1-1/10

a=9+9/10=9+0.9=9.9

b=98/11<98/10=9.8<9.9.

vậy a>b

Ta có: a=1+1/2+1+1/6+1+1/12+...+1+1/90=9+1/2+1/6+...+1/90 > 9>99/11> b. Vậy, a>b