\(P=\sqrt{x^2-4x+4}+\sqrt{x^2+6x+9}\\ =\sqrt{\left(x-2\right)^2}+\sqrt{\left(x+3\right)^2}\\ =\left|x-2\right|+\left|x+3\right|\\ =\left|2-x\right|+\left|x+3\right|\)
Áp dụng tính chất |a| + |b| ≥ |a+b| ta có:
\(P=\left|2-x\right|+\left|x+3\right|\ge\left|2-x+x+3\right|=5\)
Vậy min P = 5 khi
\(\left(2-x\right)\left(x+3\right)\ge0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2-x\ge0\\x+3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2-x\le0\\x+3\le0\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le2\\x\ge-3\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge2\\x\le-3\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow-3\le x\le2\)
