ta có:
\(x^2-6x+10=\left(x-3\right)^2+1\ge1\)
\(4x^2-24x+45=\left(2x-6\right)^2+9\ge9\)
\(\Rightarrow Q\ge\sqrt{1}+\sqrt{9}=1+3=4\)
dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x-3=0\\2x-6=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x=3\end{matrix}\right.\) (thỏa mãn)
vậy min Q = 4 khi x = 3