Đáp án:

$\begin{array}{l}
a)\left[ \begin{array}{l}
x \ge \sqrt 5 \\
x \le  - \sqrt 5 
\end{array} \right.\\
b)\left[ \begin{array}{l}
x \ge 2\sqrt 2 \\
x \le  - 2\sqrt 2 
\end{array} \right.\\
c)\left[ \begin{array}{l}
x \ge \sqrt {15} \\
x \le -\sqrt {15} 
\end{array} \right.\\
d)\left[ \begin{array}{l}
x \ge \sqrt 6 \\
x \le  - \sqrt 6 
\end{array} \right.\\
e)\left[ \begin{array}{l}
x \ge 2\\
x \le 1
\end{array} \right.\\
f)\left[ \begin{array}{l}
x \ge 3\\
x \le 2
\end{array} \right.\\
g)\left[ \begin{array}{l}
x \ge 4\\
x \le  - 3
\end{array} \right.\\
h)\left[ \begin{array}{l}
x \ge 6\\
x \le  - 5
\end{array} \right.\\
i)\left[ \begin{array}{l}
x \ge \dfrac{4}{3}\\
x \le 1
\end{array} \right.\\
j)\left[ \begin{array}{l}
x \ge 1\\
x \le \dfrac{1}{2}
\end{array} \right.
\end{array}$

Giải thích các bước giải:

$\begin{array}{l}
a)A = \sqrt {{x^2} - 5} \text{có nghĩa}\\
 \Leftrightarrow {x^2} - 5 \ge 0\\
 \Leftrightarrow {x^2} \ge 5\\
 \Leftrightarrow \left[ \begin{array}{l}
x \ge \sqrt 5 \\
x \le  - \sqrt 5 
\end{array} \right.\\
\text{Vậy} \left[ \begin{array}{l}
x \ge \sqrt 5 \\
x \le  - \sqrt 5 
\end{array} \right.\\
b)B = \sqrt {{x^2} - 8} \text{có nghĩa}\\
 \Leftrightarrow {x^2} - 8 \ge 0\\
 \Leftrightarrow {x^2} \ge 8\\
 \Leftrightarrow \left[ \begin{array}{l}
x \ge 2\sqrt 2 \\
x \le  - 2\sqrt 2 
\end{array} \right.\\
\text{Vậy} \left[ \begin{array}{l}
x \ge 2\sqrt 2 \\
x \le  - 2\sqrt 2 
\end{array} \right.\\
c)C = \sqrt {{x^2} - 15} \text{có nghĩa}\\
 \Leftrightarrow {x^2} - 15 \ge 0\\
 \Leftrightarrow {x^2} \ge 15\\
 \Leftrightarrow \left[ \begin{array}{l}
x \ge \sqrt {15} \\
x \le -\sqrt {15} 
\end{array} \right.\\
\text{Vậy} \left[ \begin{array}{l}
x \ge \sqrt {15} \\
x \le -\sqrt {15} 
\end{array} \right.\\
d)D = \sqrt {{x^2} - 6} \text{có nghĩa}\\
 \Leftrightarrow {x^2} - 6 \ge 0\\
 \Leftrightarrow {x^2} \ge 6\\
 \Leftrightarrow \left[ \begin{array}{l}
x \ge \sqrt 6 \\
x \le  - \sqrt 6 
\end{array} \right.\\
\text{Vậy} \left[ \begin{array}{l}
x \ge \sqrt 6 \\
x \le  - \sqrt 6 
\end{array} \right.\\
e)E = \sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} \text{có nghĩa}\\
 \Leftrightarrow \left( {x - 1} \right)\left( {x - 2} \right) \ge 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x \ge 2\\
x \le 1
\end{array} \right.\\
\text{Vậy} \left[ \begin{array}{l}
x \ge 2\\
x \le 1
\end{array} \right.\\
f)F = \sqrt {\left( {x - 2} \right)\left( {x - 3} \right)} \text{có nghĩa}\\
 \Leftrightarrow \left( {x - 2} \right)\left( {x - 3} \right) \ge 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x \ge 3\\
x \le 2
\end{array} \right.\\
\text{Vậy} \left[ \begin{array}{l}
x \ge 3\\
x \le 2
\end{array} \right.\\
g)G = \sqrt {{x^2} - x - 12} \text{có nghĩa}\\
 \Leftrightarrow {x^2} - x - 12 \ge 0\\
 \Leftrightarrow \left( {x - 4} \right)\left( {x + 3} \right) \ge 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x \ge 4\\
x \le  - 3
\end{array} \right.\\
\text{Vậy} \left[ \begin{array}{l}
x \ge 4\\
x \le  - 3
\end{array} \right.\\
h)H = \sqrt {{x^2} - x - 30} \text{có nghĩa}\\
 \Leftrightarrow {x^2} - x - 30 \ge 0\\
 \Leftrightarrow \left( {x - 6} \right)\left( {x + 5} \right) \ge 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x \ge 6\\
x \le  - 5
\end{array} \right.\\
\text{Vậy} \left[ \begin{array}{l}
x \ge 6\\
x \le  - 5
\end{array} \right.\\
i)I = \sqrt {3{x^2} - 7x + 4} \text{có nghĩa}\\
 \Leftrightarrow 3{x^2} - 7x + 4 \ge 0\\
 \Leftrightarrow \left( {x - 1} \right)\left( {3x - 4} \right) \ge 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x \ge \dfrac{4}{3}\\
x \le 1
\end{array} \right.\\
\text{Vậy}\left[ \begin{array}{l}
x \ge \dfrac{4}{3}\\
x \le 1
\end{array} \right.\\
j)J = \sqrt {4{x^2} - 6x + 2} \text{có nghĩa}\\
 \Leftrightarrow 4{x^2} - 6x + 2 \ge 0\\
 \Leftrightarrow 2\left( {x - 1} \right)\left( {2x - 1} \right) \ge 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x \ge 1\\
x \le \dfrac{1}{2}
\end{array} \right.\\
\text{Vậy} \left[ \begin{array}{l}
x \ge 1\\
x \le \dfrac{1}{2}
\end{array} \right.
\end{array}$