Đáp án:
$\begin{array}{l}
3)Dkxd:x \ge - 3\\
\sqrt {{x^2} + 4x + 5} = x + 3\\
\Leftrightarrow {x^2} + 4x + 5 = {x^2} + 6x + 9\\
\Leftrightarrow 2x = - 4\\
\Leftrightarrow x = - 2\left( {tmdk} \right)\\
Vậy\,x = - 2\\
4)Dkxd:\left\{ \begin{array}{l}
8 + 5x \ge 0\\
9 - 3x \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge - \dfrac{8}{5}\\
x \le 3
\end{array} \right.\\
\Leftrightarrow \dfrac{{ - 8}}{5} \le x \le 3\\
\sqrt {8 + 5x} = \sqrt {9 - 3x} \\
\Leftrightarrow 8 + 5x = 9 - 3x\\
\Leftrightarrow 8x = 1\\
\Leftrightarrow x = \dfrac{1}{8}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{1}{8}\\
5)Dkxd:x \le 2\\
\sqrt {8 - 4x} + 2\sqrt {18 - 9x} - \dfrac{3}{5}\sqrt {50 - 25x} = 15\\
\Leftrightarrow 2\sqrt {2 - x} + 2.3\sqrt {2 - x} - \dfrac{3}{5}.5\sqrt {2 - x} = 15\\
\Leftrightarrow 2\sqrt {2 - x} + 6\sqrt {2 - x} - 3\sqrt {2 - x} = 15\\
\Leftrightarrow 5\sqrt {2 - x} = 15\\
\Leftrightarrow \sqrt {2 - x} = 3\\
\Leftrightarrow 2 - x = 9\\
\Leftrightarrow x = - 7\left( {tmdk} \right)\\
Vậy\,x = - 7
\end{array}$
