Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0\\
3 + 2\sqrt x = 5\\
\Rightarrow 2\sqrt x = 2\\
\Rightarrow \sqrt x = 1\\
\Rightarrow x = 1\left( {tmdk} \right)\\
b)\sqrt {{x^2} - 10 + 25} = 5\\
\Rightarrow \sqrt {{{\left( {x - 5} \right)}^2}} = 5\\
\Rightarrow \left| {x - 5} \right| = 5\\
\Rightarrow \left[ \begin{array}{l}
x - 5 = 5\\
x - 5 = - 5
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 10\\
x = 0
\end{array} \right.\\
c)\sqrt {{x^2} - 12} = x + 3\left( {dkxd:x \ge - 3} \right)\\
\Rightarrow {x^2} - 12 = {\left( {x + 3} \right)^2}\\
\Rightarrow {x^2} - 12 = {x^2} + 6x + 9\\
\Rightarrow 6x = - 21\\
\Rightarrow x = \dfrac{{ - 7}}{2}\left( {tmdk} \right)\\
d)x - 9\sqrt x + 14 = 0\left( {dkxd:x \ge 0} \right)\\
\Rightarrow {\left( {\sqrt x } \right)^2} - 2\sqrt x - 7\sqrt x + 14 = 0\\
\Rightarrow \left( {\sqrt x - 2} \right)\left( {\sqrt x - 7} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x = 2\\
\sqrt x = 7
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 4\left( {tmdk} \right)\\
x = 49\left( {tmdk} \right)
\end{array} \right.\\
e)\sqrt {{x^2} + 6x + 9} = x - 1\left( {dkxd:x \ge 1} \right)\\
\Rightarrow {x^2} + 6x + 9 = {x^2} - 2x + 1\\
\Rightarrow 8x = - 8\\
\Rightarrow x = - 1\left( {ktm} \right)\\
\Rightarrow pt\,vô\,nghiệm\\
f)\sqrt {x + \sqrt {2x - 1} } + \sqrt {x - \sqrt {2x - 1} } = \sqrt 2 \\
\left( {Dkxd:x \ge \dfrac{1}{2}} \right)\\
\Rightarrow x + \sqrt {2x - 1} + 2\sqrt {x + \sqrt {2x - 1} } .\sqrt {x - \sqrt {2x - 1} } \\
+ x - \sqrt {2x - 1} = 2\\
\Rightarrow 2x + 2\sqrt {{x^2} - \left( {2x - 1} \right)} = 2\\
\Rightarrow \sqrt {{x^2} - 2x + 1} = 1 - x\left( {dkxd:x \le 1} \right)\\
\Rightarrow \left| {x - 1} \right| = 1 - x\\
\Rightarrow x - 1 \le 0\\
\Rightarrow x \le 1\\
Vay\,\dfrac{1}{2} \le x \le 1
\end{array}$
