Ta có : \(|x-10|+|x-5|=|x-10|+|5-x|\ge|x-10+5-x|=|-5|=5\)
\(\Rightarrow minA=5\Leftrightarrow\left(x-10\right)\left(5-x\right)\ge0\)
\(TH1:\hept{\begin{cases}x-10\ge0\\5-x\ge0\end{cases}\Rightarrow}\hept{\begin{cases}x\ge10\\5\ge x\end{cases}}\Rightarrow\hept{\begin{cases}x\ge10\\x\le5\end{cases}\Rightarrow}10\le x\le5\)(vô lý)
\(TH2:\hept{\begin{cases}x-10< 0\\5-x< 0\end{cases}\Rightarrow}\hept{\begin{cases}x< 10\\5>x\end{cases}\Rightarrow\hept{\begin{cases}x< 10\\x>5\end{cases}\Rightarrow}5< x< 10}\)(thoả mãn)
Vậy \(minA=5\Leftrightarrow5< x< 10\)
\(A=\left|x-10\right|+\left|x-5\right|=\left|x-10\right|+\left|-x+5\right|\ge\left|x-10-x+5\right|=\left|-5\right|=5\)
dấu = xảy ra khi \(\left(x-10\right).\left(-x+5\right)\ge0\)
\(\Rightarrow5\le x\le10\)
Vậy min A=10 khi và chỉ khi \(5\le x\le10\)