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 >

2 >

8²=(2³)²=2⁶

\(2^6< 8^2\)

\(5^3< 3^5\)

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

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

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

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

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

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

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

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

<=>Tự làm

2/7 < 4/9

5/7 > 15/26

9/13 < 3/5

6/11 < 3/5

3/7 > 5/11

7/11 > 12/17

13/18 > 15/23

13/6 < 17/9

2/7 < 4/9

57 < 15/26

9/13 > 3/5

6/11 >3/5

7/11 < 12/17

13/18 < 15/23

13/6 < 17/9