Đáp án: $\left[ \begin{array}{l}x\geq 2\\x\leq 1\end{array} \right.$
Giải thích các bước giải:
$|x^2-3x+2|=x^2-3x+2\\\Leftrightarrow x^{2}-3x+2\geq 0\\\Leftrightarrow x^{2}-x-2x+2\geq 0\\\Leftrightarrow x(x-1)-2(x-1)\geq 0\\\Leftrightarrow (x-1)(x-2)\geq 0\\\Leftrightarrow \left[ \begin{array}{l}\left\{\begin{matrix}
x-1\geq 0\\
x-2\geq 0
\end{matrix}\right.\\\left\{\begin{matrix}
x-1\leq 0\\
x-2\leq 0
\end{matrix}\right.\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}\left\{\begin{matrix}
x\geq 1\\
x\geq 2
\end{matrix}\right.\\\left\{\begin{matrix}
x\leq 1\\
x\leq 2
\end{matrix}\right.\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x\geq 2\\x\leq 1\end{array} \right.$
