Đáp án:
\(x \in \left[ {3; + \infty } \right) \cup \left\{ {1;2} \right\}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:{x^2} - 3x + 2 \ge 0\\
\to \left( {x - 2} \right)\left( {x - 1} \right) \ge 0\\
\to x \in \left( { - \infty ;1} \right] \cup \left[ {2; + \infty } \right)\\
Xét:\left\{ \begin{array}{l}
x - 3 = 0\\
{x^2} - 3x + 2 = 0
\end{array} \right. \to \left\{ \begin{array}{l}
x = 3\\
x = 2\\
x = 1
\end{array} \right.
\end{array}\)
BXD:
x -∞ 1 2 3 +∞
f(x) - 0 + 0 - 0 +
\(\begin{array}{l}
\to x \in \left[ {1;2} \right] \cup \left[ {3; + \infty } \right)\\
KL:x \in \left[ {3; + \infty } \right) \cup \left\{ {1;2} \right\}
\end{array}\)
