a: \(\dfrac{5+2\sqrt{5}}{\sqrt{5}+\sqrt{2}}=\dfrac{\left(5+2\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)}{3}=\dfrac{5\sqrt{5}-5\sqrt{2}+10-2\sqrt{10}}{3}\)

b: \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\sqrt{\left(2-\sqrt{3}\right)^2}=2-\sqrt{3}\)

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

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

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

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 \(\sqrt{2}-\sqrt{6}>1-\sqrt{3}\)

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 

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

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

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

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

\(x^2-\left(\sqrt{3}+\sqrt{5}\right).x+\sqrt{3}.\sqrt{5}=0\)

\(\Leftrightarrow x^2-\sqrt{3}.x-\sqrt{5}.x+\sqrt{3}.\sqrt{5}=0\)

\(\Leftrightarrow x^2-\sqrt{3}.x-\sqrt{5}.x+\sqrt{3}.\sqrt{5}=0\)

\(\Leftrightarrow x\left(x-\sqrt{3}\right)-\sqrt{5}\left(x-\sqrt{3}\right)=0\)

\(\Leftrightarrow\left(x-\sqrt{5}\right)\left(x-\sqrt{3}\right)=0\)

\(\Leftrightarrow\int^{x-\sqrt{5}=0}_{x-\sqrt{3}=0}\Leftrightarrow\int^{x=\sqrt{5}}_{x=\sqrt{3}}\)

Vậy x \(\in\left\{\sqrt{3};\sqrt{5}\right\}\)

\(<=>x^2-\sqrt{3}x-\sqrt{5}x+\sqrt{15}=0<=>x\left(x-\sqrt{3}\right)-\sqrt{5}\left(x-\sqrt{3}\right)=0<=>\left(x-\sqrt{3}\right)\left(x-\sqrt{5}\right)=0\)

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