Đáp án:

$\begin{array}{l}
6)Dkxd:\left\{ \begin{array}{l}
2x + 1 \ge 0\\
x - 3 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - \dfrac{1}{2}\\
x \ge 3
\end{array} \right. \Leftrightarrow x \ge 3\\
\sqrt {2x + 1}  = 2 + \sqrt {x - 3} \\
 \Leftrightarrow 2x + 1 = 4 + 4\sqrt {x - 3}  + x - 3\\
 \Leftrightarrow x = 4\sqrt {x - 3} \\
 \Leftrightarrow {x^2} = 16\left( {x - 3} \right)\\
 \Leftrightarrow {x^2} - 16x + 48 = 0\\
 \Leftrightarrow {x^2} - 4x - 12x + 48 = 0\\
 \Leftrightarrow \left( {x - 4} \right)\left( {x - 12} \right) = 0\\
 \Leftrightarrow x = 4;x = 12\left( {tmdk} \right)\\
Vậy\,x = 4;x = 12\\
7)Dkxd:\left\{ \begin{array}{l}
x + 4 \ge 0\\
1 - x \ge 0\\
1 - 2x \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - 4\\
x \le 1\\
x \le \dfrac{1}{2}
\end{array} \right. \Leftrightarrow  - 4 \le x \le \dfrac{1}{2}\\
\sqrt {x + 4}  - \sqrt {1 - x}  = \sqrt {1 - 2x} \\
 \Leftrightarrow \sqrt {x + 4}  = \sqrt {1 - x}  + \sqrt {1 - 2x} \\
 \Leftrightarrow x + 4 = 1 - x + 2\sqrt {1 - x} .\sqrt {1 - 2x}  + 1 - 2x\\
 \Leftrightarrow 4x + 2 = 2\sqrt {2{x^2} - 3x + 1} \\
 \Leftrightarrow 2x + 1 = \sqrt {2{x^2} - 3x + 1} \\
 \Leftrightarrow 4{x^2} + 4x + 1 = 2{x^2} - 3x + 1\\
 \Leftrightarrow 2{x^2} + 7x = 0\\
 \Leftrightarrow x\left( {2x + 7} \right) = 0\\
 \Leftrightarrow x = 0;x =  - \dfrac{7}{2}\left( {tmdk} \right)\\
Vậy\,x = 0;x =  - \dfrac{7}{2}\\
8)Dkxd:\left\{ \begin{array}{l}
3x + 4 \ge 0\\
2x + 1 \ge 0\\
x + 3 \ge 0
\end{array} \right. \Leftrightarrow x \ge \dfrac{{ - 1}}{2}\\
\sqrt {3x + 4}  - \sqrt {2x + 1}  = \sqrt {x + 3} \\
 \Leftrightarrow \sqrt {3x + 4}  = \sqrt {2x + 1}  + \sqrt {x + 3} \\
 \Leftrightarrow 3x + 4 = 2x + 1 + 2\sqrt {2x + 1} .\sqrt {x + 3}  + x + 3\\
 \Leftrightarrow 2\sqrt {2x + 1} .\sqrt {x + 3}  = 0\\
 \Leftrightarrow x =  - \dfrac{1}{2}\left( {tm} \right);x =  - 3\left( {ktm} \right)\\
Vậy\,x =  - \dfrac{1}{2}
\end{array}$