Ta có:
x2 +4x + 4= (x + 2)2
X2 - 4x + 4 = (x - 2)2
Suy ra, ta có Q = x + 2 + x - 2 = 2x
\(Q=\sqrt{x^2+4x+4}+\sqrt{x^2-4x+4}\)
\(=\sqrt{\left(x+2\right)^2}+\sqrt{\left(x-2\right)^2}\)
\(=|x+2|+|x-2|\)
\(=|x+2|+|2-x|\ge|x+2+2-x|=4\)
\(\Rightarrow Q_{min}=4\)\(\Leftrightarrow\left(x+2\right)\left(2-x\right)\ge0\)
Th1 : \(\hept{\begin{cases}x+2\ge0\\2-x\ge0\end{cases}\Rightarrow\hept{\begin{cases}x\ge-2\\x\le2\end{cases}\Rightarrow-2\le x\le}2}\)
Th2 : \(\hept{\begin{cases}x+2< 0\\2-x< 0\end{cases}\Rightarrow\hept{\begin{cases}x< -2\\x>2\end{cases}\Rightarrow}x\in\varnothing}\)
Vậy \(Q_{min}=4\Leftrightarrow-2\le x\le2\)
