Đáp án:
$\begin{array}{l}
1)Dkxd:\\
a) - 9x + 3 \ge 0\\
\Rightarrow 9x \le 3\\
\Rightarrow x \le \dfrac{1}{3}\\
b)\sqrt {\dfrac{2}{{3x + 2}}} \\
\Rightarrow \dfrac{2}{{3x + 2}} \ge 0\\
\Rightarrow 3x + 2 > 0\\
\Rightarrow x > - \dfrac{2}{3}\\
c)\left\{ \begin{array}{l}
x - 7 \ge 0\\
x - 3 \ge 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ge 7\\
x \ge 3
\end{array} \right. \Rightarrow x \ge 7\\
d)\left\{ \begin{array}{l}
x + 3 \ge 0\\
x + 2 \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ge - 3\\
x \ne - 2
\end{array} \right.\\
B2)\\
1)\sqrt {{{\left( {1 + \sqrt 3 } \right)}^2}} + \sqrt {{{\left( {1 - \sqrt 3 } \right)}^2}} \\
= 1 + \sqrt 3 + \sqrt 3 - 1\\
= 2\sqrt 3 \\
2)\sqrt {3 + 2\sqrt 2 } + \sqrt {18 - 8\sqrt 2 } \\
= \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} + \sqrt {{{\left( {4 - \sqrt 2 } \right)}^2}} \\
= \sqrt 2 + 1 + 4 - \sqrt 2 \\
= 5\\
3)\sqrt {4 - 2\sqrt 3 } + \sqrt {4 + 2\sqrt 3 } \\
= \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} \\
= \sqrt 3 - 1 + \sqrt 3 + 1\\
= 2\sqrt 3 \\
4)\sqrt {7 + 4\sqrt 3 } + \sqrt {7 - 4\sqrt 3 } \\
= \sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} + \sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} \\
= 2 + \sqrt 3 + 2 - \sqrt 3 \\
= 4\\
B3)\\
a)\sqrt {3x - 9} = 4\left( {dkxd:x \ge 3} \right)\\
\Rightarrow 3x - 9 = 16\\
\Rightarrow 3x = 25\\
\Rightarrow x = \dfrac{{25}}{3}\left( {tmdk} \right)\\
b)\sqrt {2x + 8} - 3 = 0\left( {dk:x \ge - 4} \right)\\
\Rightarrow \sqrt {2x + 8} = 3\\
\Rightarrow 2x + 8 = 9\\
\Rightarrow 2x = 1\\
\Rightarrow x = \dfrac{1}{2}\left( {tmdk} \right)\\
c)\sqrt {{x^2} + 4} = x - 2\left( {dkxd:x \ge 2} \right)\\
\Rightarrow {x^2} + 4 = {x^2} - 4x + 4\\
\Rightarrow x = 0\left( {ktm} \right)\\
\Rightarrow pt\,\text{vô}\,\text{nghiệm}\\
d)\sqrt {{x^2} - 10x + 25} = 3 - 19x\left( {dk:x \le \dfrac{3}{{19}}} \right)\\
\Rightarrow \sqrt {{{\left( {x - 5} \right)}^2}} = 3 - 19x\\
\Rightarrow \left| {x - 5} \right| = 3 - 19x\\
\Rightarrow \left[ \begin{array}{l}
x - 5 = 3 - 19x\\
x - 5 = 19x - 3
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{8}{{20}} = \dfrac{2}{5}\left( {ktm} \right)\\
x = \dfrac{{ - 1}}{9}\left( {tm} \right)
\end{array} \right.\\
\text{Vậy}\,x = - \dfrac{1}{9}
\end{array}$
