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 \(\sqrt{2}-\sqrt{6}>1-\sqrt{3}\)

\(\left(2-\sqrt{3}\right)^x+\left(7-4\sqrt{3}\right)\left(2+\sqrt{3}\right)^x=4\left(2-\sqrt{3}\right)\)

Ta có: \(2-\sqrt{3}=\frac{1}{2+\sqrt{3}}\)

\(7-4\sqrt{3}=\left(2+\sqrt{3}\right)^2\)

\(\left(2-\sqrt{3}\right)^x+\left(7-4\sqrt{3}\right)\left(2+\sqrt{3}\right)^x=4\left(2-\sqrt{3}\right)\)

<=> \(\frac{1}{\left(2+\sqrt{3}\right)^x}+\left(2-\sqrt{3}\right)^2\left(2+\sqrt{3}\right)^x=4\left(2-\sqrt{3}\right)\)

<=> \(1+\left(2-\sqrt{3}\right)^2\left(2+\sqrt{3}\right)^x\left(2+\sqrt{3}\right)^x=4\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)^x\)

<=> \(1+\left(2-\sqrt{3}\right)^2\left(2+\sqrt{3}\right)^{2x}=4\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)^x\)

Đặt:  \(\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)^x=t\)

Ta có pt ẩn t: \(1+t^2=4t\)

<=> \(t^2-4t+1=0\Leftrightarrow\orbr{\begin{cases}t=2-\sqrt{3}\\t=2+\sqrt{3}\end{cases}}\)

+) Với \(t=2+\sqrt{3}\), ta có: 

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

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

<=> x=2 

Trường hợp còn lại em làm tương tự

\(\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}=\dfrac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\dfrac{2-\sqrt{3}}{4-3}+\dfrac{2+\sqrt{3}}{4-3}=2-\sqrt{3}+2+\sqrt{3}=4\)

1)

\(5x^2-\sqrt{3}x-1=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{23}-\sqrt{3}}{10}\\x=\frac{\sqrt{23}+\sqrt{3}}{10}\end{cases}}\)

Ta có: \(\dfrac{7-4\sqrt{3}}{\sqrt{3}-2}-\dfrac{28-10\sqrt{3}}{5-\sqrt{3}}\)

\(=\dfrac{\left(\sqrt{3}-2\right)^2}{\sqrt{3}-2}-\dfrac{\left(5-\sqrt{3}\right)^2}{5-\sqrt{3}}\)

\(=\sqrt{3}-2-5+\sqrt{3}\)

=-7

a)

ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

Ta có: \(\dfrac{2x}{x-3}=\dfrac{x^2+11x-6}{x^2-9}\)

\(\Leftrightarrow\dfrac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+11x-6}{\left(x-3\right)\left(x+3\right)}\)

Suy ra: \(2x^2+6x=x^2+11x-6\)

\(\Leftrightarrow2x^2+6x-x^2-11x+6=0\)

\(\Leftrightarrow x^2-5x+6=0\)

\(\Leftrightarrow x^2-2x-3x+6=0\)

\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)

Vậy: S={2}

b) Ta có: \(3x^2+\left(1-\sqrt{3}\right)x+\sqrt{3}-4=0\)

\(\Leftrightarrow3x^2-\left(\sqrt{3}-1\right)x+\sqrt{3}-4=0\)

\(\Leftrightarrow3x^2-\left(\sqrt{3}-1\right)x+\sqrt{3}-1-3=0\)

\(\Leftrightarrow\left(3x^2-3\right)-\left(\sqrt{3}-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)-\left(\sqrt{3}-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x+3-\sqrt{3}+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x+4-\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x+4-\sqrt{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\3x=\sqrt{3}-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{\sqrt{3}-4}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{1;\dfrac{\sqrt{3}-4}{3}\right\}\)