Đáp án:

\(\begin{array}{l}
a,\\
x = \dfrac{8}{3}\\
b,\\
\left[ \begin{array}{l}
x = 4\\
x =  - 4
\end{array} \right.\\
c,\\
\left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x = \dfrac{1}{2}
\end{array} \right.\\
d,\\
Phương\,\,trình\,\,vô\,\,nghiệm
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
DKXD:\,\,3x - 4 \ge 0 \Leftrightarrow x \ge \dfrac{4}{3}\\
\sqrt {3x - 4}  = 2\\
 \Leftrightarrow 3x - 4 = {2^2}\\
 \Leftrightarrow 3x - 4 = 4\\
 \Leftrightarrow 3x = 8\\
 \Leftrightarrow x = \dfrac{8}{3}\\
b,\\
DKXD:\,\,\,9{x^2} \ge 0,\,\,\,\forall x\\
\sqrt {9{x^2}}  = \left| { - 12} \right|\\
 \Leftrightarrow \sqrt {{{\left( {3x} \right)}^2}}  = 12\\
 \Leftrightarrow \left| {3x} \right| = 12\\
 \Leftrightarrow \left[ \begin{array}{l}
3x = 12\\
3x =  - 12
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x =  - 4
\end{array} \right.\\
c,\\
DKXD:\,\,4{x^2} - 12x + 9 \ge 0,\,\,\,\forall x\\
\sqrt {4{x^2} - 12x + 9}  = 2\\
 \Leftrightarrow \sqrt {{{\left( {2x} \right)}^2} - 2.2x.3 + {3^2}}  = 2\\
 \Leftrightarrow \sqrt {{{\left( {2x - 3} \right)}^2}}  = 2\\
 \Leftrightarrow \left| {2x - 3} \right| = 2\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - 3 = 2\\
2x - 3 =  - 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = 5\\
2x = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x = \dfrac{1}{2}
\end{array} \right.\\
d,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
2x - 6 \ge 0\\
2 - x \ge 0\\
\sqrt {2 - x}  \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2x - 6 \ge 0\\
2 - x > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2x \ge 3\\
2 > x
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 3\\
x < 2
\end{array} \right.\\
 \Rightarrow Phương\,\,trình\,\,vô\,\,nghiệm
\end{array}\)