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