Đáp án:

$\begin{array}{l}
a)\sqrt 3 {x^2} - \left( {1 - \sqrt 3 } \right)x - 1 = 0\\
 \Rightarrow \Delta  = {\left( {1 - \sqrt 3 } \right)^2} - 4.\sqrt 3 .\left( { - 1} \right)\\
 = 1 - 2\sqrt 3  + 3 + 4\sqrt 3 \\
 = 1 + 2\sqrt 3  + 3\\
 = {\left( {1 + \sqrt 3 } \right)^2}\\
 \Rightarrow \sqrt \Delta   = 1 + \sqrt 3 \\
 \Rightarrow \left\{ \begin{array}{l}
{x_1} = \dfrac{{1 - \sqrt 3  + \left( {1 + \sqrt 3 } \right)}}{{2\sqrt 3 }} = \dfrac{1}{{\sqrt 3 }} = \dfrac{{\sqrt 3 }}{3}\\
{x_2} = \dfrac{{1 - \sqrt 3  - \left( {1 + \sqrt 3 } \right)}}{{2\sqrt 3 }} =  - 1
\end{array} \right.\\
b){x^2} - \left( {2 + \sqrt 3 } \right)x + 2\sqrt 3  = 0\\
 \Rightarrow \Delta  = {\left( {2 + \sqrt 3 } \right)^2} - 4.2\sqrt 3 \\
 = 4 + 4\sqrt 3  + 3 - 8\sqrt 3 \\
 = 4 - 4\sqrt 3  + 3\\
 = {\left( {2 - \sqrt 3 } \right)^2}\\
 \Rightarrow \sqrt \Delta   = 2 - \sqrt 3 \\
 \Rightarrow \left\{ \begin{array}{l}
{x_1} = \dfrac{{2 + \sqrt 3  + \left( {2 - \sqrt 3 } \right)}}{2} = 2\\
{x_1} = \dfrac{{2 + \sqrt 3  - \left( {2 - \sqrt 3 } \right)}}{2} = \sqrt 3 
\end{array} \right.\\
c)\sqrt 6 {x^2} - 4\sqrt 5  + 8 = 0\\
 \Rightarrow {x^2} = \dfrac{{4\sqrt 5  - 8}}{{\sqrt 6 }} = \dfrac{{2\sqrt {30}  - 4\sqrt 6 }}{3}\\
 \Rightarrow x =  \pm 0,62\\
d){x^2} + 2\left( {1 + \sqrt 5 } \right)x + 1 + 2\sqrt 5  = 0\\
 \Rightarrow \Delta ' = {\left( {1 + \sqrt 5 } \right)^2} - 1 - 2\sqrt 5 \\
 = 1 + 2\sqrt 5  + 5 - 1 - 2\sqrt 5 \\
 = 5\\
 \Rightarrow \sqrt {\Delta '}  = \sqrt 5 \\
 \Rightarrow \left\{ \begin{array}{l}
{x_1} = \dfrac{{1 + \sqrt 5  + \sqrt 5 }}{1} = 2\sqrt 5  + 1\\
{x_2} = 1
\end{array} \right.
\end{array}$

$\begin{array}{l} a)\sqrt 3 {x^2} - \left( {1 - \sqrt 3 } \right)x - 1 = 0\\  \Rightarrow \Delta  = {\left( {1 - \sqrt 3 } \right)^2} - 4.\sqrt 3 .\left( { - 1} \right)\\  = 1 - 2\sqrt 3  + 3 + 4\sqrt 3 \\  = 1 + 2\sqrt 3  + 3\\  = {\left( {1 + \sqrt 3 } \right)^2}\\  \Rightarrow \sqrt \Delta   = 1 + \sqrt 3 \\  \Rightarrow \left\{ \begin{array}{l} {x_1} = \dfrac{{1 - \sqrt 3  + \left( {1 + \sqrt 3 } \right)}}{{2\sqrt 3 }} = \dfrac{1}{{\sqrt 3 }} = \dfrac{{\sqrt 3 }}{3}\\ {x_2} = \dfrac{{1 - \sqrt 3  - \left( {1 + \sqrt 3 } \right)}}{{2\sqrt 3 }} =  - 1 \end{array} \right.\\ b){x^2} - \left( {2 + \sqrt 3 } \right)x + 2\sqrt 3  = 0\\  \Rightarrow \Delta  = {\left( {2 + \sqrt 3 } \right)^2} - 4.2\sqrt 3 \\  = 4 + 4\sqrt 3  + 3 - 8\sqrt 3 \\  = 4 - 4\sqrt 3  + 3\\  = {\left( {2 - \sqrt 3 } \right)^2}\\  \Rightarrow \sqrt \Delta   = 2 - \sqrt 3 \\  \Rightarrow \left\{ \begin{array}{l} {x_1} = \dfrac{{2 + \sqrt 3  + \left( {2 - \sqrt 3 } \right)}}{2} = 2\\ {x_1} = \dfrac{{2 + \sqrt 3  - \left( {2 - \sqrt 3 } \right)}}{2} = \sqrt 3  \end{array} \right.\\ c)\sqrt 6 {x^2} - 4\sqrt 5  + 8 = 0\\  \Rightarrow {x^2} = \dfrac{{4\sqrt 5  - 8}}{{\sqrt 6 }} = \dfrac{{2\sqrt {30}  - 4\sqrt 6 }}{3}\\  \Rightarrow x =  \pm 0,62\\ d){x^2} + 2\left( {1 + \sqrt 5 } \right)x + 1 + 2\sqrt 5  = 0\\  \Rightarrow \Delta ' = {\left( {1 + \sqrt 5 } \right)^2} - 1 - 2\sqrt 5 \\  = 1 + 2\sqrt 5  + 5 - 1 - 2\sqrt 5 \\  = 5\\  \Rightarrow \sqrt {\Delta '}  = \sqrt 5 \\  \Rightarrow \left\{ \begin{array}{l} {x_1} = \dfrac{{1 + \sqrt 5  + \sqrt 5 }}{1} = 2\sqrt 5  + 1\\ {x_2} = 1 \end{array} \right. \end{array}$