$\sqrt{x^{2} - 3x + 2} \geq x - 5$
TH1:
$\left\{\begin{matrix}x^{2} - 3x + 2 \geq 0\\ x - 5 \leq 0\end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}\left ( x - 1 \right )\left ( x - 2 \right ) \geq 0\\ x \leq 5\end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}\left[ \begin{array}{l}x \leq 1\\x \geq 2\end{array} \right.\\ x \leq 5\end{matrix}\right.$
$\Leftrightarrow \left[ \begin{array}{l}x \leq 1\\2 \leq x \leq 5\end{array} \right.$
TH2:
$\left\{\begin{matrix}x - 5 \geq 0\\ x^{2} - 3x + 2 \geq \left ( x - 5 \right )^{2} \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}x \geq 5\\ x^{2} - 3x + 2 \geq x^{2} - 10x + 25 \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}x \geq 5\\ 7x \geq 23 \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}x \geq 5\\ x \geq \dfrac{23}{7} \end{matrix}\right.$
$\Leftrightarrow x \geq 5$
Kết hợp hai TH ta được $x \leq 1; x \geq 2$
ĐS $x \leq 1; x \geq 2$
