Đáp án:
$\begin{array}{l}
a)y = \sqrt {3x - 2} + \sqrt {3 - 2x} \\
Dkxd:\left\{ \begin{array}{l}
3x - 2 \ge 0\\
3 - 2x \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{2}{3}\\
x \le \dfrac{3}{2}
\end{array} \right.\\
\Leftrightarrow \dfrac{2}{3} \le x \le \dfrac{3}{2}\\
Vậy\,TXD:D = \left[ {\dfrac{2}{3};\dfrac{3}{2}} \right]\\
b)y = \dfrac{2}{{\sqrt {3 - 2x} }}\\
Dkxd:3 - 2x > 0\\
\Leftrightarrow x < \dfrac{3}{2}\\
Vậy\,TXD:D = \left( { - \infty ;\dfrac{3}{2}} \right)\\
c)y = \dfrac{{\sqrt {8 - x} }}{x}\\
Dkxd:\left\{ \begin{array}{l}
8 - x \ge 0\\
x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le 8\\
x \ne 0
\end{array} \right.\\
Vậy\,TXD:D = \left( { - \infty ;8} \right]\backslash \left\{ 0 \right\}\\
d)Dkxd:\left\{ \begin{array}{l}
2x + 1 \ge 0\\
x - 2 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge - \dfrac{1}{2}\\
x \ne 2
\end{array} \right.\\
Vậy\,TXD:D = \left[ { - \dfrac{1}{2}; + \infty } \right)\backslash \left\{ 2 \right\}\\
e)Dkxd:\left\{ \begin{array}{l}
x \ne 0\\
2x \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne 0\\
x \ge 0
\end{array} \right. \Leftrightarrow x > 0\\
VậyTXD:D = \left( {0; + \infty } \right)\\
f)Dkxd:\left\{ \begin{array}{l}
{x^2} - 3 \ne 0\\
{x^2} \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne \sqrt 3 ;x \ne - \sqrt 3 \\
x \ne 0
\end{array} \right.\\
Vậy\,TXD:D = R\backslash \left\{ { - \sqrt 3 ;0;\sqrt 3 } \right\}\\
g)Dkxd:{x^2} - 8x + 15 > 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {x - 5} \right) > 0\\
\Leftrightarrow \left[ \begin{array}{l}
x > 5\\
x < 3
\end{array} \right.\\
Vậy\,TXD:D = R\backslash \left[ {3;5} \right]\\
h)Dkxd:\left| x \right| \ge 2\\
\Leftrightarrow \left[ \begin{array}{l}
x \ge 2\\
x \le - 2
\end{array} \right.\\
Vậy\,TXD:D = R\backslash \left( { - 2;2} \right)\\
i)DKxd:\left| {x - 2} \right| \ge 4\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 \ge 4 \Leftrightarrow x \ge 6\\
x - 2 \le - 4 \Leftrightarrow x \le - 2
\end{array} \right.\\
Vậy\,TXD:D = R\backslash \left( { - 2;6} \right)\\
j)Dkxd: - \left| {x + 1} \right| - 3 \ge 0\\
\Leftrightarrow \left| {x + 1} \right| \le - 3\left( {ktm} \right)\\
Vậy\,x \in \emptyset
\end{array}$
