Đáp án:
k. \(1 \ge x \ge - 2\)
Giải thích các bước giải:
\(\begin{array}{l}
n.DK:{x^2} - 5x + 6 \ge 0\\
\to \left( {x - 3} \right)\left( {x - 2} \right) \ge 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 3 \ge 0\\
x - 2 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 3 \le 0\\
x - 2 \le 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
x \ge 3\\
x \le 2
\end{array} \right.\\
l.DK:3 - {x^2} \ge 0\\
\to 3 \ge {x^2}\\
\to \sqrt 3 \ge \left| x \right|\\
\to \sqrt 3 \ge x \ge - \sqrt 3 \\
k.DK: - {x^2} - x + 2 \ge 0\\
\to \left( {1 - x} \right)\left( {x + 2} \right) \ge 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
1 - x \ge 0\\
x + 2 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
1 - x \le 0\\
x + 2 \le 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
1 \ge x \ge - 2\\
\left\{ \begin{array}{l}
x \ge 1\\
x \le - 2
\end{array} \right.\left( l \right)
\end{array} \right.\\
h.DK:9{x^2} - 6x + 1 \ge 0\\
\to {\left( {3x - 1} \right)^2} \ge 0\left( {ld} \right)\forall x \in R
\end{array}\)
⇒ ĐIều kiện với mọi x
