\(M=\sqrt{x^2+6x+9}+\sqrt{x^2-4x+4}\)
\(M=\sqrt{\left(x+3\right)^2}+\sqrt{\left(x-2\right)^2}\)
\(M=\left|x+3\right|+\left|x-2\right|\)
\(M=\left|x+3\right|+\left|2-x\right|\ge\left|x+3+2-x\right|=\left|5\right|=5\)
Dấu "=" xảy ra \(\Leftrightarrow-3\le x\le2\)