Đáp án:

\(\begin{array}{l}
1,\,\,\,\,\left\{ \begin{array}{l}
x \ne \sqrt 6 \\
x \ne  - \sqrt 6 
\end{array} \right.\\
2,\,\,\,\,\,\forall x\\
3,\,\,\,\,\,x \ge 0\\
4,\,\,\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x \ne 4\\
x \ne 9
\end{array} \right.\\
5,\,\,\,\,\left\{ \begin{array}{l}
x \ge 1\\
x \ne 2
\end{array} \right.\\
6,\,\,\,\,2 \le x \le 5\\
7,\,\,\,\, - 3 < x \le 1\\
8,\,\,\,\,x \ne 2\\
9,\,\,\,\, - 2 < x < 1
\end{array}\)

Giải thích các bước giải:

 Các hàm số đã cho xác định khi và chỉ khi:

\(\begin{array}{l}
1,\\
6 - {x^2} \ne 0 \Leftrightarrow {x^2} \ne 6 \Leftrightarrow \left\{ \begin{array}{l}
x \ne \sqrt 6 \\
x \ne  - \sqrt 6 
\end{array} \right.\\
2,\\
{x^2} - 2x + 3 \ge 0 \Leftrightarrow \left( {{x^2} - 2x + 1} \right) + 2 \ge 0\\
 \Leftrightarrow \left( {{x^2} - 2.x.1 + {1^2}} \right) + 2 \ge 0 \Leftrightarrow {\left( {x - 1} \right)^2} + 2 \ge 0,\,\,\,\forall x\\
3,\\
\left\{ \begin{array}{l}
x \ge 0\\
\sqrt x  + 1 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\sqrt x  \ne  - 1
\end{array} \right. \Leftrightarrow x \ge 0\\
4,\\
\left\{ \begin{array}{l}
x \ge 0\\
x - 5\sqrt x  + 6 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\left( {x - 2\sqrt x } \right) + \left( { - 3\sqrt x  + 6} \right) \ne 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\sqrt x \left( {\sqrt x  - 2} \right) - 3\left( {\sqrt x  - 2} \right) \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\left( {\sqrt x  - 2} \right)\left( {\sqrt x  - 3} \right) \ne 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\sqrt x  - 2 \ne 0\\
\sqrt x  - 3 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
\sqrt x  \ne 2\\
\sqrt x  \ne 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 4\\
x \ne 9
\end{array} \right.\\
5,\\
\left\{ \begin{array}{l}
x \ge 0\\
x - 1 \ge 0\\
\sqrt {x - 1}  - 1 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ge 1\\
\sqrt {x - 1}  \ne 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
x - 1 \ne 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
x \ne 2
\end{array} \right.\\
6,\\
\left\{ \begin{array}{l}
x - 2 \ge 0\\
5 - x \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 2\\
5 \ge x
\end{array} \right. \Leftrightarrow 2 \le x \le 5\\
7,\\
\left\{ \begin{array}{l}
x + 3 \ge 0\\
\sqrt {x + 3}  \ne 0\\
1 - x \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x + 3 > 0\\
1 - x \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x >  - 3\\
x \le 1
\end{array} \right. \Leftrightarrow  - 3 < x \le 1\\
8,\\
\left\{ \begin{array}{l}
4 - 4x + {x^2} \ge 0\\
\sqrt {4 - 4x + {x^2}}  \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{2^2} - 2.2.x + {x^2} \ge 0\\
\sqrt {{2^2} - 2.2.x + {x^2}}  \ne 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {2 - x} \right)^2} \ge 0\\
\sqrt {{{\left( {2 - x} \right)}^2}}  \ne 0
\end{array} \right. \Leftrightarrow 2 - x \ne 0 \Leftrightarrow x \ne 2\\
9,\\
\left\{ \begin{array}{l}
2 - x - {x^2} \ge 0\\
\sqrt {2 - x - {x^2}}  \ne 0
\end{array} \right. \Leftrightarrow 2 - x - {x^2} > 0 \Leftrightarrow  - \left( {{x^2} + x - 2} \right) > 0\\
 \Leftrightarrow {x^2} + x - 2 < 0\\
 \Leftrightarrow \left( {{x^2} - x} \right) + \left( {2x - 2} \right) < 0\\
 \Leftrightarrow x\left( {x - 1} \right) + 2\left( {x - 1} \right) < 0\\
 \Leftrightarrow \left( {x + 2} \right)\left( {x - 1} \right) < 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 2 > 0\\
x - 1 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 2 < 0\\
x - 1 > 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x >  - 2\\
x < 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x <  - 2\\
x > 1
\end{array} \right.
\end{array} \right. \Leftrightarrow  - 2 < x < 1
\end{array}\)