Ta có |x - 15| + |x - 1| = 14
<=> |15 - x| + |x - 1| = 14
mà |15 - x| + |x - 1| \(\le\left|15-x+x-1\right|=14\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}15-x\ge0\\x-1\ge0\end{cases}\Rightarrow\hept{\begin{cases}x\le15\\x\ge1\end{cases}\Rightarrow}1\le x\le15}\)
Vậy \(1\le x\le15\)