Đáp án:

\(\begin{array}{l}
a,\,\,\,\,\,x = 2\\
b,\,\,\,\,\,\,x = 2\\
c,\,\,\,\,\,\,x = \dfrac{3}{4}\\
d,\,\,\,\,\,x = 5\\
e,\,\,\,\,\,\,x = \dfrac{8}{3}\\
g,\,\,\,\,\,x =  - 3
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
DKXD:\,\,\,4x + 1 \ge 0 \Leftrightarrow x \ge  - \dfrac{1}{4}\\
\sqrt {4x + 1}  = 3\\
 \Leftrightarrow {\sqrt {4x + 1} ^2} = {3^2}\\
 \Leftrightarrow 4x + 1 = 9\\
 \Leftrightarrow 4x = 8\\
 \Leftrightarrow x = 2\\
b,\\
DKXD:\,\,\,\,x \ge  - 2\\
\dfrac{1}{2}\sqrt {4x + 8}  + \sqrt {36x + 72}  = 16 - \sqrt {x + 2} \\
 \Leftrightarrow \dfrac{1}{2}.\sqrt {4.\left( {x + 2} \right)}  + \sqrt {36.\left( {x + 2} \right)}  = 16 - \sqrt {x + 2} \\
 \Leftrightarrow \dfrac{1}{2}.\sqrt 4 .\sqrt {x + 2}  + \sqrt {36} .\sqrt {x + 2}  + \sqrt {x + 2}  = 16\\
 \Leftrightarrow \dfrac{1}{2}.2\sqrt {x + 2}  + 6\sqrt {x + 2}  + \sqrt {x + 2}  = 16\\
 \Leftrightarrow \sqrt {x + 2}  + 6\sqrt {x + 2}  + \sqrt {x + 2}  = 16\\
 \Leftrightarrow 8\sqrt {x + 2}  = 16\\
 \Leftrightarrow \sqrt {x + 2}  = 2\\
 \Leftrightarrow x + 2 = 4\\
 \Leftrightarrow x = 2\\
c,\\
DKXD:\,\,\,\forall x\\
\sqrt {{x^2} - 4x + 4}  + 1 = 3x\\
 \Leftrightarrow \sqrt {{x^2} - 2.x.2 + {2^2}}  + 1 = 3x\\
 \Leftrightarrow \sqrt {{{\left( {x - 2} \right)}^2}}  = 3x - 1\\
 \Leftrightarrow \left| {x - 2} \right| = 3x - 1\\
 \Leftrightarrow \left\{ \begin{array}{l}
3x - 1 \ge 0\\
\left[ \begin{array}{l}
x - 2 = 3x - 1\\
x - 2 = 1 - 3x
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{3}\\
\left[ \begin{array}{l}
x - 3x =  - 1 + 2\\
x + 3x = 1 + 2
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{3}\\
\left[ \begin{array}{l}
 - 2x = 1\\
4x = 3
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{1}{3}\\
\left[ \begin{array}{l}
x =  - \dfrac{1}{2}\\
x = \dfrac{3}{4}
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow x = \dfrac{3}{4}\\
d,\\
DKXD:\,\,\,\,x \ge \dfrac{1}{2}\\
\sqrt {2x - 1}  + 1 = 4\\
 \Leftrightarrow \sqrt {2x - 1}  = 3\\
 \Leftrightarrow {\sqrt {2x - 1} ^2} = {3^2}\\
 \Leftrightarrow 2x - 1 = 9\\
 \Leftrightarrow 2x = 10\\
 \Leftrightarrow x = 5\\
e,\\
DKXD:\,\,\,\,\,x \ge  - \dfrac{1}{3}\\
3\sqrt {12x + 4}  - \dfrac{8}{3}\sqrt {27x + 9}  + \sqrt {48x + 16}  = 6\\
 \Leftrightarrow 3\sqrt {4.\left( {3x + 1} \right)}  - \dfrac{8}{3}\sqrt {9\left( {3x + 1} \right)}  + \sqrt {16.\left( {3x + 1} \right)}  = 6\\
 \Leftrightarrow 3.\sqrt 4 .\sqrt {3x + 1}  - \dfrac{8}{3}.\sqrt 9 .\sqrt {3x + 1}  + \sqrt {16} .\sqrt {3x + 1}  = 6\\
 \Leftrightarrow 3.2\sqrt {3x + 1}  - \dfrac{8}{3}.3\sqrt {3x + 1}  + 4\sqrt {3x + 1}  = 6\\
 \Leftrightarrow 6\sqrt {3x + 1}  - 8\sqrt {3x + 1}  + 4\sqrt {3x + 1}  = 6\\
 \Leftrightarrow 2\sqrt {3x + 1}  = 6\\
 \Leftrightarrow \sqrt {3x + 1}  = 3\\
 \Leftrightarrow {\sqrt {3x + 1} ^2} = {3^2}\\
 \Leftrightarrow 3x + 1 = 9\\
 \Leftrightarrow 3x = 8\\
 \Leftrightarrow x = \dfrac{8}{3}\\
g,\\
DKXD:\,\,\,\,\left\{ \begin{array}{l}
{x^2} - 9 \ge 0\\
x + 3 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left( {x - 3} \right)\left( {x + 3} \right) \ge 0\\
x \ge  - 3
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 3\\
x \le  - 3
\end{array} \right.\\
x \ge  - 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x \ge 3\\
x =  - 3
\end{array} \right.\\
\sqrt {{x^2} - 9}  + \sqrt {x + 3}  = 0\\
 \Leftrightarrow \sqrt {\left( {x - 3} \right)\left( {x + 3} \right)}  + \sqrt {x + 3}  = 0\\
 \Leftrightarrow \sqrt {x + 3} .\left( {\sqrt {x - 3}  + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt {x + 3}  = 0\\
\sqrt {x - 3}  + 1 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
\sqrt {x - 3}  =  - 1\,\,\,\,\,\,\,\left( {L,\,\,\,\sqrt {x - 3}  \ge 0,\,\,\,\forall x \ge 3} \right)
\end{array} \right.\\
 \Leftrightarrow x =  - 3
\end{array}\)