Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0\\
3\sqrt x - 1 = 8\\
\Leftrightarrow 3\sqrt x = 9\\
\Leftrightarrow \sqrt x = 3\\
\Leftrightarrow x = 9\left( {tmdk} \right)\\
Vậy\,x = 9\\
b)\sqrt {{x^2} - 6x + 9} = \left| 2 \right|\\
\Leftrightarrow {x^2} - 6x + 9 = 4\\
\Leftrightarrow {x^2} - 6x + 5 = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {x - 5} \right) = 0\\
\Leftrightarrow x = 1;x = 5\\
Vậy\,x = 1;x = 5\\
c)Dkxd:x \ge 3\\
2\sqrt {9\left( {x - 3} \right)} - \dfrac{1}{5}\sqrt {25\left( {x - 3} \right)} + \dfrac{1}{7}\sqrt {49\left( {x - 3} \right)} = 20\\
\Leftrightarrow 2.3\sqrt {x - 3} - \dfrac{1}{5}.5\sqrt {x - 3} + \dfrac{1}{7}.7\sqrt {x - 3} = 20\\
\Leftrightarrow 6\sqrt {x - 3} - \sqrt {x - 3} + \sqrt {x - 3} = 20\\
\Leftrightarrow 6\sqrt {x - 3} = 20\\
\Leftrightarrow \sqrt {x - 3} = \dfrac{{10}}{3}\\
\Leftrightarrow x - 3 = \dfrac{{100}}{9}\\
\Leftrightarrow x = \dfrac{{127}}{9}\\
Vậy\,x = \dfrac{{127}}{9}\\
d)2x + \sqrt {2x + 19} = 1\\
\Leftrightarrow \sqrt {2x + 19} = 1 - 2x\\
Dkxd:\left\{ \begin{array}{l}
2x + 19 \ge 0\\
1 - 2x \ge 0
\end{array} \right. \Leftrightarrow - \dfrac{{19}}{2} \le x \le \dfrac{1}{2}\\
\Leftrightarrow 2x + 19 = 4{x^2} - 4x + 1\\
\Leftrightarrow 4{x^2} - 6x - 18 = 0\\
\Leftrightarrow 2{x^2} - 3x - 9 = 0\\
\Leftrightarrow 2{x^2} - 6x + 3x - 9 = 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {2x + 3} \right) = 0\\
\Leftrightarrow x = \dfrac{{ - 3}}{2}\left( {do:\dfrac{{ - 19}}{2} \le x \le \dfrac{1}{2}} \right)\\
Vậy\,x = \dfrac{{ - 3}}{2}
\end{array}$
