Đáp án:
$\begin{array}{l}
1)Dkxd:x \ge 1\\
\sqrt {4x - 4} + \dfrac{1}{3}\sqrt {\dfrac{{9x - 9}}{{16}}} = \dfrac{3}{4}\\
\Leftrightarrow 2\sqrt {x - 1} + \dfrac{1}{3}.\dfrac{{3\sqrt {x - 1} }}{4} = \dfrac{3}{4}\\
\Leftrightarrow 2\sqrt {x - 1} + \dfrac{1}{4}\sqrt {x - 1} = \dfrac{3}{4}\\
\Leftrightarrow \dfrac{9}{4}\sqrt {x - 1} = \dfrac{3}{4}\\
\Leftrightarrow \sqrt {x - 1} = \dfrac{1}{3}\\
\Leftrightarrow x - 1 = \dfrac{1}{9}\\
\Leftrightarrow x = \dfrac{{10}}{9}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{{10}}{9}\\
2)Dkxd:x \ge 2\\
\sqrt {{x^2} - 2x + 1} - 2\sqrt {\dfrac{{x - 2}}{4}} = 3\\
\Leftrightarrow \sqrt {{{\left( {x - 1} \right)}^2}} - \dfrac{{2\sqrt {x - 2} }}{2} = 3\\
\Leftrightarrow x - 1 - \sqrt {x - 2} = 3\\
\Leftrightarrow x - 4 = \sqrt {x - 2} \left( {dk:x \ge 4} \right)\\
\Leftrightarrow {x^2} - 8x + 16 = x - 2\\
\Leftrightarrow {x^2} - 9x + 18 = 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {x - 6} \right) = 0\\
\Leftrightarrow x = 6\left( {do:x \ge 4} \right)\\
Vậy\,x = 6\\
3)x - \sqrt {3x + 1} = 3\\
\Leftrightarrow x - 3 = \sqrt {3x + 1} \left( {dk:x \ge 3} \right)\\
\Leftrightarrow {x^2} - 6x + 9 - 3x - 1 = 0\\
\Leftrightarrow {x^2} - 9x + 8 = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {x - 8} \right) = 0\\
\Leftrightarrow x = 8\left( {do:x \ge 3} \right)\\
Vậy\,x = 8\\
4)Dkxd:\left\{ \begin{array}{l}
x \ge - 2\\
4 - {x^2} \ge 0
\end{array} \right. \Leftrightarrow - 2 \le x \le 2\\
\sqrt {4 - {x^2}} - 2\sqrt {x + 2} = 0\\
\Leftrightarrow \sqrt {x + 2} .\left( {\sqrt {2 - x} - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 2 = 0\\
\sqrt {2 - x} = 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 2\left( {tm} \right)\\
2 - x = 4 \Leftrightarrow x = - 2\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = - 2
\end{array}$
