a) \(\left|x-10\right|+\left|x-3\right|\)
Đặt \(B=\left|x-10\right|+\left|x-3\right|\)
\(\Rightarrow B=\left|x-10\right|+\left|3-x\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(B=\left|x-10\right|+\left|3-x\right|\ge\left|x-10+3-x\right|\)
\(\Rightarrow B\ge\left|-7\right|\)
\(\Rightarrow B\ge7.\)
Dấu '' = '' xảy ra khi:
\(\left(x-10\right).\left(3-x\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-10\ge0\\3-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-10\le0\\3-x\le0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge10\\x\ge3\end{matrix}\right.\\\left\{{}\begin{matrix}x\le10\\x\le3\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3\le x\le10\\x\in\varnothing\end{matrix}\right.\)
Vậy \(MIN_B=7\) khi \(3\le x\le10.\)
Chúc bạn học tốt!
