a/ | x + 4 | < 3
=> \(\left|x+4\right|\in\left\{0;1;2\right\}\)
=> \(x+4\in\left\{0;1;-1;2;-2\right\}\)
=> \(x\in\left\{-4;-3;-5;-2;-6\right\}\)
b/ | x - 14 + 17 | + | y + 10 - 12 | ≤ 0
*Trường hợp 1: | x - 14 + 17 | + | y + 10 - 12 | < 0
=> Vô lí.
*Trường hợp 2: | x - 14 + 17 | + | y + 10 - 12 | = 0
Ta có: \(\left|x-14+17\right|\ge0\) ; \(\left|y+10-12\right|\ge0\)
=> \(\left|x-14+17\right|+\left|y+10-12\right|\ge0\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}\left|x-14+17\right|=0\\\left|y+10-12\right|=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x-14+17=0\\y+10-12=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x+3=0\\y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0-3=-3\\y=0+2=2\end{matrix}\right.\)
Vậy: x = -3; y = 2
