Để $\sqrt{x^2+3x+2}+\sqrt{2x+1}$ xác định
$\Leftrightarrow\begin{cases}x^2+3x+2\ge0\\2x+1\ge0\end{cases}$
$\Leftrightarrow\begin{cases}(x+1)(x+2)\ge0\\2x\ge-1\end{cases}$
$\Leftrightarrow\begin{cases}\left[\begin{matrix}x+1\ge0\\x+2\le0\end{matrix}\right.\\x\ge-\dfrac{1}{2}\end{cases}$
$\Leftrightarrow\begin{cases}\left[\begin{matrix}x\ge-1\\x\le-2\end{matrix}\right.\\x\ge-\dfrac{1}{2}\end{cases}$
$\Leftrightarrow x\ge-\dfrac{1}{2}$
Vậy $ĐKXĐ: x\ge-\dfrac{1}{2}$
