a) \(\left(\sqrt{\dfrac{9}{20}}-\sqrt{\dfrac{1}{2}}\right).\sqrt{2}=\sqrt{\dfrac{9}{20}.2}-\sqrt{\dfrac{1}{2}.2}=\sqrt{\dfrac{9}{10}}-1=\dfrac{3}{\sqrt{10}}-1\)

\(=\dfrac{3\sqrt{10}}{10}-1\)

b) \(\left(\sqrt{12}+\sqrt{27}-\sqrt{3}\right)\sqrt{3}=\sqrt{12.3}+\sqrt{27.3}-\sqrt{3.3}\)

\(=\sqrt{36}+\sqrt{81}-\sqrt{9}=6+9-3=12\)

c) \(\left(\sqrt{\dfrac{8}{3}}-\sqrt{24}+\sqrt{\dfrac{50}{3}}\right)\sqrt{6}=\sqrt{\dfrac{8}{3}.6}-\sqrt{24.6}+\sqrt{\dfrac{50}{3}.6}\)

\(=\sqrt{16}-\sqrt{144}+\sqrt{100}=4-12+10=2\)

\(\sqrt{3}-\frac{5}{2}>\sqrt{3}-4\text{ vì }-\frac{5}{2}>-4\)

\(\Rightarrow2.\left(\sqrt{3}-\frac{5}{2}\right)>\sqrt{3}-4\)

\(\Rightarrow2.\sqrt{3}-5>\sqrt{3}-4\)

b) vì \(\sqrt{5}-\sqrt{12}< 0\), ta có: 

 \(5\sqrt{5}-2\sqrt{3}=4\sqrt{5}+\sqrt{5}-\sqrt{12}< 4\sqrt{5}< 4\sqrt{5}+6\) 

Vậy \(5\sqrt{5}-2\sqrt{3}< 6+4\sqrt{5}\)

c)\(\sqrt{2}-\sqrt{6}=\sqrt{2}.\left(\sqrt{1}-\sqrt{3}\right)>\left(1-\sqrt{3}\right)\)

Vậy \(\sqrt{2}-\sqrt{6}>1-\sqrt{3}\)

\(2>\sqrt{3}\)

trả lời 

> 

chúc bn 

hc tốt

2 >\(\sqrt{3}\)

Chúc bn hk tt:v

#JM#

a,Ta có :  \(1-\sqrt{3}\); \(\sqrt{2}-\sqrt{6}=\sqrt{2}\left(1-\sqrt{3}\right)\Rightarrow1-\sqrt{3}< \sqrt{2}\left(1-\sqrt{3}\right)\)

Vậy \(1-\sqrt{3}< \sqrt{2}-\sqrt{6}\)

b, Đặt A =  \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)(*)

\(\sqrt{2}A=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)

\(=\sqrt{7}+1-\sqrt{7}+1-2=0\Rightarrow A=0\)

Vậy (*) = 0 

1: 

Ta có: \(\sqrt{2}-\sqrt{6}\)

\(=\sqrt{2}\left(1-\sqrt{3}\right)< 0\)

\(\Leftrightarrow1-\sqrt{3}< \sqrt{2}-\sqrt{6}\)

2:
Ta có: \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)

\(=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2}{\sqrt{2}}\)

\(=\dfrac{\sqrt{7}+1-\sqrt{7}+1-2}{\sqrt{2}}\)

=0

1. Ta có 4=2 căn 4 

Căn 4<căn 5

=> 2 căn 5 >4

2. Ta có 3^2=9 =16-7=16-căn 49

( căn 15 -1)^2

= 15 -2 căn 15 +1= 16-2 căn 15 =16- căn 60

Căn 60>căn49

=> 3> căn 15 -1

3. Ta có  6^2=36=27+9= 27+ căn 81

    (căn 26 +1)^2=26 +2 căn 26 +1=27+ 2 căn 26 =27+ căn 52

 Căn 52< căn 81 

=> 6> căn 26+1

4. Ta có (căn 2 -2)^2 =2- 4 căn 2+4=6- 4 căn 2

             (căn 3 -3 )^2 = 3 -6 căn 3 +9= 12- 6 căn 3

      Lại có 8 căn 2 =căn 128

                6 căn 3 =căn 108

=> (căn 3 -3)^2> 2(căn 2 -2)^2 

=> căn 3 -3 > căn 2-2 

\(2\sqrt{5}>4\)

\(3< \sqrt{15-1}\)

\(6>\sqrt{26-1}\)

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

\(P=\left|5-\sqrt{24}\right|+3+\sqrt{24}=5-\sqrt{24}+3+\sqrt{24}=8\)

\(P=\left|5-\sqrt{24}\right|+3+\sqrt{24}\)

TH1 : \(\left|5-\sqrt{24}\right|=5-\sqrt{24}\)

TH2 : \(\left|5-\sqrt{24}\right|=-5+\sqrt{24}\)

Với TH1 thì : \(\left|5-\sqrt{24}\right|+3+\sqrt{24}=5-\sqrt{24}+3+\sqrt{24}=8\)

Với TH2 thì : \(\left|5-\sqrt{24}\right|+3+\sqrt{24}=-5+\sqrt{24}+3+\sqrt{24}=-2+2\sqrt{24}\)

\(P=\left|5-\sqrt{24}\right|+3+\sqrt{24}\)

\(=5-\sqrt{24}+3+\sqrt{24}\)

\(=\left(5+3\right)+\left(-\sqrt{24}+\sqrt{24}\right)\)

\(=8+0=8\)

\(-3\sqrt{3}=-\sqrt{27}\)

\(-2\sqrt{7}=-\sqrt{28}\)

mà 27<28

nên \(-3\sqrt{3}>-2\sqrt{7}\)