Giải thích các bước giải:
Các biểu thức đã cho có nghĩa khi và chỉ khi:
\(\begin{array}{l}
1,\\
9{x^2} - 6x + 1 \ge 0\\
\Leftrightarrow {\left( {3x} \right)^2} - 2.3x.1 + {1^2} \ge 0\\
\Leftrightarrow {\left( {3x - 1} \right)^2} \ge 0,\,\,\,\,\forall x\\
\Rightarrow DKXD:\,\,\,\,\,\forall x\\
2,\\
6 - x - {x^2} \ge 0\\
\Leftrightarrow - \left( {{x^2} + x - 6} \right) \ge 0\\
\Leftrightarrow {x^2} + x - 6 \le 0\\
\Leftrightarrow \left( {{x^2} + 3x} \right) - \left( {2x + 6} \right) \le 0\\
\Leftrightarrow x\left( {x + 3} \right) - 2.\left( {x + 3} \right) \le 0\\
\Leftrightarrow \left( {x + 3} \right)\left( {x - 2} \right) \le 0\\
\Leftrightarrow - 3 \le x \le 2\\
\Rightarrow DKXD:\,\,\,\, - 3 \le x \le 2\\
3,\\
\left\{ \begin{array}{l}
4x - 1 \ge 0\\
2x - \sqrt {4x - 1} \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
4x \ge 1\\
4x - 2.\sqrt {4x - 1} \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge \frac{1}{4}\\
\left( {4x - 1} \right) - 2.\sqrt {4x - 1} + 1 \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge \frac{1}{4}\\
{\left( {\sqrt {4x - 1} - 1} \right)^2} \ge 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge \frac{1}{4}\\
\forall x
\end{array} \right.\\
\Rightarrow x \ge \frac{1}{4}\\
\Rightarrow DKXD:\,\,\,\,\,x \ge \frac{1}{4}
\end{array}\)
