Đáp án:
$\begin{array}{l}
a)\sqrt {\dfrac{{ - 6}}{{4 + x}}} = 5\\
\Leftrightarrow \dfrac{{ - 6}}{{4 + x}} = 25\\
\Leftrightarrow 4 + x = - \dfrac{6}{{25}}\\
\Leftrightarrow x = - \dfrac{{106}}{{25}}\\
Vậy\,x = - \dfrac{{106}}{{25}}\\
b)Dkxd:\left\{ \begin{array}{l}
x \ge 5\\
1 - x \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 5\\
x \le 1
\end{array} \right.\left( {ktm} \right)\\
Vậy\,x \in \emptyset \\
c)\sqrt {3{x^2} - 4x + 3} = 1 - 2x\left( {dk:x \le \dfrac{1}{2}} \right)\\
\Leftrightarrow 3{x^2} - 4x + 3 = 1 - 4x + 4{x^2}\\
\Leftrightarrow {x^2} = 2\\
\Leftrightarrow x = - \sqrt 2 \left( {do:x \le \dfrac{1}{2}} \right)\\
Vậy\,x = - \sqrt 2 \\
d)Dkxd:x \ge - 1\\
\sqrt {9x + 9} + \sqrt {4x + 4} = \sqrt {x + 1} \\
\Leftrightarrow 3\sqrt {x + 1} + 2\sqrt {x + 1} - \sqrt {x + 1} = 0\\
\Leftrightarrow 4\sqrt {x + 1} = 0\\
\Leftrightarrow x = - 1\left( {tmdk} \right)\\
Vậy\,x = - 1
\end{array}$
