Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 3\\
\sqrt {{x^2} - 6} = x - 3\\
\Leftrightarrow {x^2} - 6 = {\left( {x - 3} \right)^2}\\
\Leftrightarrow {x^2} - 6 = {x^2} - 6x + 9\\
\Leftrightarrow 6x = 15\\
\Leftrightarrow x = \dfrac{5}{2}\\
\text{Vậy}\,x = \dfrac{5}{2}\\
b)\sqrt {4{x^2} - 4x} = 4 - 2x\\
Dkxd:\left\{ \begin{array}{l}
4x\left( {x - 1} \right) \ge 0\\
4 - 2x \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 1\\
x \le 0
\end{array} \right.\\
x \le 2
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
1 \le x \le 2\\
x \le 0
\end{array} \right.\\
Pt \Leftrightarrow 4{x^2} - 4x = 16 - 16x + 4{x^2}\\
\Leftrightarrow 12x = 16\\
\Leftrightarrow x = \dfrac{4}{3}\left( {tmdk} \right)\\
\text{Vậy}\,x = \dfrac{4}{3}\\
c)\sqrt {{x^2} + 16} = 3x + 4\\
\left( {dkxd:x \ge \dfrac{{ - 4}}{3}} \right)\\
\Leftrightarrow {x^2} + 16 = 9{x^2} + 24x + 16\\
\Leftrightarrow 8{x^2} + 24x = 0\\
\Leftrightarrow 8x\left( {x + 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x + 3 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\left( {tm} \right)\\
x = - 3\left( {ktm} \right)
\end{array} \right.\\
\text{Vậy}\,x = 0\\
d)Dkxd:x \le \dfrac{3}{2}\\
\sqrt {{x^2} + x + 9} = 3 - 2x\\
\Leftrightarrow {x^2} + x + 9 = 9 - 12x + 4{x^2}\\
\Leftrightarrow 3{x^2} - 13x = 0\\
\Leftrightarrow x\left( {3x - 13} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\left( {tm} \right)\\
x = \dfrac{{13}}{3}\left( {ktm} \right)
\end{array} \right.\\
\text{Vậy}\,x = 0\\
e)\sqrt {4x + 1} = 2x - 1\left( {dkxd:x \ge \dfrac{1}{2}} \right)\\
\Leftrightarrow 4x + 1 = 4{x^2} - 4x + 1\\
\Leftrightarrow 4{x^2} - 8x = 0\\
\Leftrightarrow 4x\left( {x - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\left( {ktm} \right)\\
x = 2\left( {tm} \right)
\end{array} \right.\\
\text{Vậy}\,x = 2\\
f)Dkxd:x \ge 2\\
\sqrt {x + 1} + \sqrt {x - 2} = 3\\
\Leftrightarrow x + 1 + 2\sqrt {x + 1} .\sqrt {x - 2} + x - 2 = 9\\
\Leftrightarrow 2\sqrt {{x^2} - x - 2} = 10 - 2x\\
\Leftrightarrow \sqrt {{x^2} - x - 2} = 5 - x\left( {dk:x \le 5} \right)\\
\Leftrightarrow {x^2} - x - 2 = {x^2} - 10x + 25\\
\Leftrightarrow 9x = 27\\
\Leftrightarrow x = 3\left( {tmdk} \right)\\
\text{Vậy}\,x = 3
\end{array}$
