Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0\\
\sqrt x + \sqrt {\frac{x}{9}} - \frac{1}{3}\sqrt {4x} = 5\\
\Rightarrow \sqrt x + \frac{{\sqrt x }}{3} - \frac{1}{3}.2\sqrt x = 5\\
\Rightarrow \sqrt x .\left( {1 + \frac{1}{3} - \frac{2}{3}} \right) = 5\\
\Rightarrow \sqrt x .\frac{2}{3} = 5\\
\Rightarrow \sqrt x = \frac{{15}}{2}\\
\Rightarrow x = \frac{{225}}{4}\left( {tm} \right)\\
\text{Vậy}\,x = \frac{{225}}{4}\\
c)Dkxd:\left\{ \begin{array}{l}
x \ge 3\\
x \le 5
\end{array} \right. \Rightarrow 3 \le x \le 5\\
\sqrt {x - 3} + \sqrt {5 - x} = 2\\
\Rightarrow x - 3 + 2\sqrt {x - 3} .\sqrt {5 - x} + 5 - x = 4\\
\Rightarrow 2 + 2\sqrt {\left( {x - 3} \right)\left( {5 - x} \right)} = 4\\
\Rightarrow \sqrt {\left( {x - 3} \right)\left( {5 - x} \right)} = 1\\
\Rightarrow 5x - {x^2} - 15 + 3x = 1\\
\Rightarrow {x^2} - 8x + 16 = 0\\
\Rightarrow {\left( {x - 4} \right)^2} = 0\\
\Rightarrow x = 4\left( {tmdk} \right)\\
\text{Vậy}\,x = 4
\end{array}$
