ĐK: $(x-1)(x-2)(x-3)\ge 0$
Đặt $f(x)=(x-1)(x-2)(x-3)$
$f(x)=0\\↔\left[\begin{array}{1}x-1=0\\x-2=0\\x-3=0\end{array}\right.\\↔\left[\begin{array}{1}x=1\\x=2\\x=3\end{array}\right.$
BXD:
$\begin{array}{|c|cc|}\hline x&-\infty&&1&&2&&3&&+\infty\\\hline x-1&&-&0&+&|&+&|&+&\\\hline x-2&&-&|&-&0&+&|&+&\\\hline x-3&&-&|&-&|&-&0&+&\\\hline f(x)&&-&0&+&0&-&0&+&\\\hline\end{array}$
Vì $f(x)\ge 0$
$→\left[\begin{array}{1}1\le x\le 2\\x\ge 3\end{array}\right.$
Vậy $1\le x\le 2$ hoặc $x\ge 3$ thì biểu thức xác định