Đáp án:
$\begin{array}{l}
a)Dkxd:x \in R\\
Vậy\,TXD:D = R\\
b)Dkxd:{x^2} - 6x + 10\# 0\\
\Leftrightarrow {x^2} - 6x + 9 + 1\# 0\\
\Leftrightarrow {\left( {x - 3} \right)^2} + 1\# 0\\
\Leftrightarrow TXD:D = R\\
c)Dkxd:\left\{ \begin{array}{l}
x - 1 \ge 0\\
\left| x \right| - 2\# 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
\left| x \right|\# 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
x\# 2
\end{array} \right.\\
Vậy\,TXD:D = \left[ {1;0} \right)\backslash \left\{ 2 \right\}\\
d)Dkxd:\left\{ \begin{array}{l}
x \ge 0\\
2 - x \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \le 2
\end{array} \right. \Leftrightarrow 0 \le x \le 2\\
Vậy\,TXD:D = \left[ {0;2} \right]\\
e)Dkxd:\left\{ \begin{array}{l}
{x^2} - 9 \ge 0\\
x - 2\# 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x^2} \ge 9\\
x\# 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x \ge 3\\
x \le - 3
\end{array} \right.\\
Vậy\,TXD:D = R\backslash \left[ { - 3;3} \right]
\end{array}$
