a, Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(A=\left|x\right|+\left|8-x\right|\ge\left|x+8-x\right|=\left|8\right|=8\)
Dấu " = " khi \(\left\{{}\begin{matrix}x\ge0\\8-x\ge0\end{matrix}\right.\Rightarrow0\le x\le8\)
Vậy \(MIN_A=8\) khi \(0\le x\le8\)
b, Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(B=\left|x-2015\right|+\left|2016-x\right|\ge\left|x-2015+2016-x\right|=\left|-1\right|=1\)
Dấu " = " khi \(\left\{{}\begin{matrix}x-2015\ge0\\2016-x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge2015\\x\le2016\end{matrix}\right.\)
Vậy \(MIN_B=1\) khi \(2015\le x\le2016\)
