Đáp án:

$\begin{array}{l}
6)Dkxd:x \ge 2\\
\sqrt {x - 1 + 2\sqrt {x - 2} }  = 2\\
 \Rightarrow \sqrt {x - 2 + 2\sqrt {x - 2}  + 1}  = 2\\
 \Rightarrow \sqrt {{{\left( {\sqrt {x - 2}  + 1} \right)}^2}}  = 2\\
 \Rightarrow \sqrt {x - 2}  + 1 = 2\\
 \Rightarrow \sqrt {x - 2}  = 1\\
 \Rightarrow x - 2 = 1\\
 \Rightarrow x = 3\left( {tmdk} \right)\\
7)Dkxd:x \ge 4\\
\sqrt {x + 4\sqrt {x - 4} }  = 2\\
 \Rightarrow \sqrt {x - 4 + 4\sqrt {x - 4}  + 4}  = 2\\
 \Rightarrow \sqrt {{{\left( {\sqrt {x - 4}  + 2} \right)}^2}}  = 2\\
 \Rightarrow \sqrt {x - 4}  + 2 = 2\\
 \Rightarrow \sqrt {x - 4}  = 0\\
 \Rightarrow x = 4\left( {tmdk} \right)\\
8)Dkxd:\left[ \begin{array}{l}
x \ge 3\\
x \le  - 3
\end{array} \right.\\
\sqrt {{x^2} - 9}  + \sqrt {{x^2} - 6x + 9}  = 0\\
 \Rightarrow \left\{ \begin{array}{l}
\sqrt {{x^2} - 9}  = 0\\
\sqrt {{x^2} - 6x + 9}  = 0
\end{array} \right.\\
 \Rightarrow x = 3\left( {tmdk} \right)\\
9)Dkxd:x \ge \frac{3}{2}\\
\sqrt {2x - 2 + 2\sqrt {2x - 3} }  + \sqrt {2x + 13 + 8\sqrt {2x - 3} }  = 5\\
 \Rightarrow \sqrt {2x - 3 + 2\sqrt {2x - 3}  + 1}  + \sqrt {2x - 3 + 8\sqrt {2x - 3}  + 16}  = 5\\
 \Rightarrow \sqrt {{{\left( {\sqrt {2x - 3}  + 1} \right)}^2}}  + \sqrt {{{\left( {\sqrt {2x - 3}  + 4} \right)}^2}}  = 5\\
 \Rightarrow \sqrt {2x - 3}  + 1 + \sqrt {2x - 3}  + 4 = 5\\
 \Rightarrow \sqrt {2x - 3}  = 0\\
 \Rightarrow 2x - 3 = 0\\
 \Rightarrow x = \frac{3}{2}\left( {tmdk} \right)\\
10)Dkxd:x \ge  - \frac{3}{2}\\
 \Rightarrow {x^2} + 4x + 5 = 2\sqrt {2x + 3} \\
 \Rightarrow {\left( {x + 2} \right)^2} + 1 = 2\sqrt {2x + 3} \left( {vn} \right)\\
11)\sqrt {x + 3}  + \sqrt {y - 2}  + \sqrt {z - 3}  = \frac{1}{2}\left( {x + y + z + 1} \right)\\
 \Rightarrow 2\sqrt {x + 3}  + 2\sqrt {y - 2}  + 2\sqrt {z - 3}  = x + y + z + 1\\
 \Rightarrow x + 3 - 2\sqrt {x + 3}  + 1 + y - 2 - 2\sqrt {y - 2}  + 1\\
 + z - 3 - 2\sqrt {z - 3}  + 1 = 0\\
 \Rightarrow {\left( {\sqrt {x + 3}  - 1} \right)^2} + {\left( {\sqrt {y - 2}  - 1} \right)^2}\\
 + {\left( {\sqrt {z - 3}  - 1} \right)^2} = 0\\
 \Rightarrow \left\{ \begin{array}{l}
\sqrt {x + 3}  = 1\\
\sqrt {y - 2}  = 1\\
\sqrt {z - 3}  = 1
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x =  - 2\\
y = 3\\
z = 4
\end{array} \right.\\
12)x + y + 4 = 2\sqrt x  + 4\sqrt {y - 1} \\
 \Rightarrow x - 2\sqrt x  + 1 + y - 1 - 4\sqrt {y - 1}  + 4 = 0\\
 \Rightarrow {\left( {\sqrt x  - 1} \right)^2} + {\left( {\sqrt {y - 1}  - 2} \right)^2} = 0\\
 \Rightarrow \left\{ \begin{array}{l}
\sqrt x  = 1\\
\sqrt {y - 1}  = 2
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x = 1\\
y - 1 = 4
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x = 1\\
y = 5
\end{array} \right.
\end{array}$