Ta có A = |x - 2015| + |x - 2016|
= |x - 2015| + |2016 - x|
\(\ge\)|x - 2015 + 2016 - x| = 1
Dấu "=" xảy ra <=> \(\left(x-2015\right)\left(2016-x\right)\ge0\)
TH1 : \(\hept{\begin{cases}x-2015\ge0\\2016-x\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x\ge2015\\x\le2016\end{cases}}\Rightarrow2015\le x\le2016\)
TH2 : \(\hept{\begin{cases}x-2015\le0\\2016-x\le0\end{cases}}\Rightarrow\hept{\begin{cases}x\le2015\\x\ge2016\end{cases}}\left(\text{loại}\right)\)
Vậy Min A = 1 <=> \(2015\le x\le2016\)
b) Ta có B = |x - 5| + |x - 7|+ |2x - 18|
= |x - 5| + |x - 7|+ |18 - 2x|
\(\ge\)|x - 5 + x - 7| + |18 - 2x|
= |2x - 12| + |18 - 2x|
\(\ge\)|2x - 12 + 18 - 2x| = 6
Dấu "=" xảy ra <=> \(\left(2x-12\right)\left(18-2x\right)\ge0\)
TH1 : \(\hept{\begin{cases}2x-12\ge0\\18-2x\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x\ge6\\x\le9\end{cases}}\Rightarrow6\le x\le9\)
TH2 : \(\hept{\begin{cases}2x-12\le0\\18-2x\le0\end{cases}}\Rightarrow\hept{\begin{cases}x\le6\\x\ge9\end{cases}}\)(loại)
Vậy Min B = 6 <=> \(6\le x\le9\)
