$\sqrt[]{x^2-6x+10}$ có nghĩa $⇔x∈R$
Do $x^2-6x+10=(x^2-6x+9)+1=(x-3)^2+1≥1$ vì $(x-3)^2≥0∀x$
$\sqrt[]{4x^2+12+10}$ có nghĩa $⇔4x^2+12x+10≥0$
$⇔4x^2+12x+12-2≥0$
$⇔4(x^2+4x+4)≥2$
$⇔(x+2)^2≥\dfrac{1}{2}$
$⇔$
\(\left[ \begin{array}{l}x+2=\dfrac{1}{\sqrt[]2}\\x+2=-\dfrac{1}{\sqrt[]2}\end{array} \right.\)$⇔$\(\left[ \begin{array}{l}x=\dfrac{1}{\sqrt[]2}-2\\x=-\dfrac{1}{\sqrt[]2}-2\end{array} \right.\)
