Để căn thức có nghĩa thì $x^2-5x+6≥0$
$⇔x^2-2x-3x+6≥0$
$⇔x(x-2)-3(x-2)≥0$
$⇔(x-3)(x-2)≥0$
⇒\(\left[ \begin{array}{l}x-3≥0\\x-2≥0\end{array} \right.\) hay \(\left[ \begin{array}{l}x-3≤0\\x-2≤0\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}x≥3\\x≤2\end{array} \right.\)