Đáp án:

$\begin{array}{l}
9)Dkxd:\left\{ \begin{array}{l}
x + 3 \ge 0\\
x - 4 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - 3\\
x \ge 4
\end{array} \right. \Leftrightarrow x \ge 4\\
\sqrt {x + 3}  - \sqrt {x - 4}  = 1\\
 \Leftrightarrow \sqrt {x + 3}  = \sqrt {x - 4}  + 1\\
 \Leftrightarrow x + 3 = x - 4 + 2\sqrt {x - 4}  + 1\\
 \Leftrightarrow 2\sqrt {x - 4}  = 6\\
 \Leftrightarrow \sqrt {x - 4}  = 3\\
 \Leftrightarrow x - 4 = 9\\
 \Leftrightarrow x = 13\left( {tmdk} \right)\\
Vậy\,x = 13
\end{array}$

$\begin{array}{l}
10)Dkxd:\left\{ \begin{array}{l}
15 - x \ge 0\\
3 - x \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le 15\\
x \le 3
\end{array} \right. \Leftrightarrow x \le 3\\
 \Leftrightarrow \sqrt {15 - x}  + \sqrt {3 - x}  = 6\\
 \Leftrightarrow 15 - x + 2\sqrt {15 - x} .\sqrt {3 - x}  + 3 - x = 36\\
 \Leftrightarrow 2\sqrt {\left( {15 - x} \right)\left( {3 - x} \right)}  = 2x - 18\\
 \Leftrightarrow \sqrt {{x^2} - 18x + 45}  = x - 9\\
 \Leftrightarrow {x^2} - 18x + 45 = {x^2} - 18x + 81\\
 \Leftrightarrow 45 = 81\left( {ktm} \right)\\
Vậy\,x \in \emptyset \\
11)Dkxd:\left\{ \begin{array}{l}
x + 3 \ge 0\\
10 - x \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - 3\\
x \le 10
\end{array} \right. \Leftrightarrow  - 3 \le x \le 10\\
\sqrt {x + 3}  + \sqrt {10 - x}  = 5\\
 \Leftrightarrow x + 3 + 2\sqrt {x + 3} .\sqrt {10 - x}  + 10 - x = 25\\
 \Leftrightarrow 2\sqrt {\left( {x + 3} \right)\left( {10 - x} \right)}  = 12\\
 \Leftrightarrow \sqrt { - {x^2} + 7x + 30}  = 6\\
 \Leftrightarrow  - {x^2} + 7x + 30 = 36\\
 \Leftrightarrow {x^2} - 7x + 6 = 0\\
 \Leftrightarrow \left( {x - 1} \right)\left( {x - 6} \right) = 0\\
 \Leftrightarrow x = 1;x = 6\left( {tmdk} \right)\\
Vậy\,x = 1;x = 6
\end{array}$