Đáp án:

\(\begin{array}{l}
f,\\
x = 5\,\,\,\,\\
g,\\
x = 1\\
h,\\
Phương\,\,trình\,\,vô \,\,nghiệm\\
i,\\
x = 2\\
j,\\
x = 5
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
f,\\
DKXD:\,\,\,2x - 1 \ge 0 \Leftrightarrow x \ge \dfrac{1}{2}\\
\sqrt {2x - 1}  = x - 2\\
 \Leftrightarrow \left\{ \begin{array}{l}
x - 2 \ge 0\\
{\sqrt {2x - 1} ^2} = {\left( {x - 2} \right)^2}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge 2\\
2x - 1 = {x^2} - 4x + 4
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge 2\\
{x^2} - 6x + 5 = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge 2\\
\left( {x - 1} \right)\left( {x - 5} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge 2\\
\left[ \begin{array}{l}
x - 1 = 0\\
x - 5 = 0
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge 2\\
\left[ \begin{array}{l}
x = 1\\
x = 5
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow x = 5\,\,\,\,\\
g,\\
DKXD:\,\,\,\,5 - x \ge 0 \Leftrightarrow x \le 5\\
\sqrt {5 - x}  = x + 1\\
 \Leftrightarrow \left\{ \begin{array}{l}
x + 1 \ge 0\\
{\sqrt {5 - x} ^2} = {\left( {x + 1} \right)^2}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - 1\\
5 - x = {x^2} + 2x + 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - 1\\
{x^2} + 3x - 4 = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - 1\\
\left( {x + 4} \right)\left( {x - 1} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - 1\\
\left[ \begin{array}{l}
x + 4 = 0\\
x - 1 = 0
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - 1\\
\left[ \begin{array}{l}
x =  - 4\\
x = 1
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow x = 1\\
h,\\
DKXD:\,\,\,4{x^2} - 12x + 9 \ge 0 \Leftrightarrow {\left( {2x - 3} \right)^2} \ge 0,\,\,\,\forall x\\
\sqrt {4{x^2} - 12x + 9}  = 2x - 5\\
 \Leftrightarrow \sqrt {{{\left( {2x} \right)}^2} - 2.2x.3 + {3^2}}  = 2x - 5\\
 \Leftrightarrow \sqrt {{{\left( {2x - 3} \right)}^2}}  = 2x - 5\\
 \Leftrightarrow \left| {2x - 3} \right| = 2x - 5\\
 \Leftrightarrow \left\{ \begin{array}{l}
2x - 5 \ge 0\\
\left[ \begin{array}{l}
2x - 3 = 2x - 5\\
2x - 3 = 5 - 2x
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{5}{2}\\
\left[ \begin{array}{l}
 - 3 =  - 5\,\,\,\left( L \right)\\
4x = 8
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{5}{2}\\
x = 2
\end{array} \right.\\
 \Rightarrow Phương\,\,trình\,\,vô\,\,nghiệm\\
i,\\
DKXD:\,\,\,x \ge  - 2\\
\sqrt {4x + 8}  - 5\sqrt {x + 2}  + 3\sqrt {9x + 18}  = 12\\
 \Leftrightarrow \sqrt {4.\left( {x + 2} \right)}  - 5\sqrt {x + 2}  + 3\sqrt {9\left( {x + 2} \right)}  = 12\\
 \Leftrightarrow \sqrt {{2^2}.\left( {x + 2} \right)}  - 5\sqrt {x + 2}  + 3\sqrt {{3^2}.\left( {x + 2} \right)}  = 12\\
 \Leftrightarrow 2\sqrt {x + 2}  - 5\sqrt {x + 2}  + 3.3\sqrt {x + 2}  = 12\\
 \Leftrightarrow 2\sqrt {x + 2}  - 5\sqrt {x + 2}  + 9\sqrt {x + 2}  = 12\\
 \Leftrightarrow 6\sqrt {x + 2}  = 12\\
 \Leftrightarrow \sqrt {x + 2}  = 2\\
 \Leftrightarrow x + 2 = 4\\
 \Leftrightarrow x = 2\\
j,\\
DKXD:\,\,\,x \ge 4\\
\sqrt {4x - 16}  - 5\sqrt {x - 4}  = 3\sqrt {9x - 36}  - 12\\
 \Leftrightarrow \sqrt {4.\left( {x - 4} \right)}  - 5\sqrt {x - 4}  = 3\sqrt {9\left( {x - 4} \right)}  - 12\\
 \Leftrightarrow \sqrt {{2^2}.\left( {x - 4} \right)}  - 5\sqrt {x - 4}  = 3\sqrt {{3^2}.\left( {x - 4} \right)}  - 12\\
 \Leftrightarrow 2\sqrt {x - 4}  - 5\sqrt {x - 4}  = 3.3\sqrt {x - 4}  - 12\\
 \Leftrightarrow  - 3\sqrt {x - 4}  = 9\sqrt {x - 4}  - 12\\
 \Leftrightarrow  - 12\sqrt {x - 4}  =  - 12\\
 \Leftrightarrow \sqrt {x - 4}  = 1\\
 \Leftrightarrow x - 4 = 1\\
 \Leftrightarrow x = 5
\end{array}\)