\(M=\left|x-3\right|+\left|x-5\right|+x^2-8x+2019\)
\(=\left|x-3\right|+\left|5-x\right|+x^2-8x+16+2013\)
\(=\left|x-3\right|+\left|5-x\right|+\left(x-4\right)^2+2013\)
Ta thấy \(\left|x-3\right|+\left|5-x\right|\ge\left|x-3+5-x\right|\ge2\)
\(\left(x-4\right)^2\ge0\)
\(\Rightarrow M=\left|x-3\right|+\left|5-x\right|+\left(x-4\right)^2+2013\ge2+0+2013=2015\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\left|x-3\right|+\left|5-x\right|=2\\\left(x-4\right)^2=0\end{cases}\Leftrightarrow}x=4\)
hicc mình trừ nhầm :">
Dòng 2 trở đi là + 2003 nhá
GTNN = 2005
T^T
