Đáp án:
$\begin{array}{l}
1)Dkxd:x \ge 1\\
\sqrt {4\left( {x - 1} \right)} = 2\\
\Leftrightarrow 2\sqrt {x - 1} = 2\\
\Leftrightarrow \sqrt {x - 1} = 1\\
\Leftrightarrow x - 1 = 1\\
\Leftrightarrow x = 2\left( {tmdk} \right)\\
\text{Vậy}\,x = 2\\
2)Dkxd:x \ge - \dfrac{1}{2}\\
3\sqrt {\dfrac{2}{9}x + \dfrac{1}{9}} - 2\sqrt {2x + 1} - 3 = 0\\
\Leftrightarrow 3.\dfrac{1}{3}.\sqrt {2x + 1} - 2\sqrt {2x + 1} - 3 = 0\\
\Leftrightarrow \sqrt {2x + 1} - 2\sqrt {2x + 1} - 3 = 0\\
\Leftrightarrow - \sqrt {2x + 1} = 3\left( {ktm} \right)\\
\text{Vậy pt vô nghiệm}\\
3)Dkxd:x \ge 1\\
\sqrt {9x - 9} - 2\sqrt {4x - 4} + \sqrt {x - 1} = 4\\
\Leftrightarrow 3\sqrt {x - 1} - 2.2.\sqrt {x - 1} + \sqrt {x - 1} = 4\\
\Leftrightarrow 0 = 4\left( {ktm} \right)\\
\text{Vậy pt vô nghiệm}\\
4)Dkxd:x \ge 3\\
\sqrt {{x^2} - 9} - \sqrt {x - 3} = 0\\
\Leftrightarrow \sqrt {{x^2} - 9} = \sqrt {x - 3} \\
\Leftrightarrow {x^2} - 9 = x - 3\\
\Leftrightarrow {x^2} - x - 6 = 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {x + 2} \right) = 0\\
\Leftrightarrow x = 3\left( {do:x \ge 3} \right)\\
\text{Vậy}\,x = 3\\
5)Dkxd:x \ge \dfrac{1}{2}\\
\sqrt {4{x^2} - 1} - 2\sqrt {2x + 1} = 0\\
\Leftrightarrow \sqrt {4{x^2} - 1} = \sqrt {8x + 4} \\
\Leftrightarrow 4{x^2} - 1 = 8x + 4\\
\Leftrightarrow 4{x^2} - 8x - 5 = 0\\
\Leftrightarrow \left( {2x - 5} \right)\left( {2x + 1} \right) = 0\\
\Leftrightarrow x = \dfrac{5}{2}\left( {do:x \ge \dfrac{1}{2}} \right)\\
\text{Vậy}\,x = \dfrac{5}{2}\\
6)Dkxd:x > - 2\\
\dfrac{{3x + 2}}{{\sqrt {x + 2} }} = 2\sqrt {x + 2} \\
\Leftrightarrow 3x + 2 = 2\left( {x + 2} \right)\\
\Leftrightarrow x = 2\left( {tmdk} \right)\\
\text{Vậy}\,x = 2\\
7)Dkxd:x \ge 2\\
\sqrt {{x^2} - 4} - 2\sqrt {x + 2} = 0\\
\Leftrightarrow \sqrt {x - 2} .\sqrt {x + 2} - 2\sqrt {x + 2} = 0\\
\Leftrightarrow \sqrt {x + 2} .\left( {\sqrt {x - 2} - 2} \right) = 0\\
\Leftrightarrow \sqrt {x - 2} = 2\left( {do:x \ge 2} \right)\\
\Leftrightarrow x - 2 = 4\\
\Leftrightarrow x = 6\left( {tmdk} \right)\\
\text{Vậy}\,x = 6
\end{array}$
