Giải thích các bước giải:
$|x+4|<3\\\text{Ta có:} |x + 4| \ge 0 \forall x\\ \text{mà:} |x+4|<3\\\Rightarrow -3 < x + 4 < 3\\ \Rightarrow -7 < x <-1\\ \Rightarrow x \in \{-6;-5;-4;-3;-2\}$
Vậy........
.
$1<|x-2|<4\\ \begin{cases} |x-2| > 1 \Leftrightarrow \left[\begin{array}{l} x-2>1\\x-2<-1\end{array}\right. \Leftrightarrow \left[\begin{array}{l} x > 3\\x<1\end{array}\right.\\|x-2|< 4 \Leftrightarrow -4 <x -2<4\\ \Leftrightarrow -2 < x < 6 \Leftrightarrow x \in \{-1;0;1;2;3;4;5\} \\\end{cases}\\ \Rightarrow x \in \{ -1;0;4;5\}$
Vậy..........
.
$|x-14+17|+|y+10-12| \le0\\ \Leftrightarrow |x+3|+|y-2| \le 0\\ \text{Ta có:} \begin{cases}|x+3| \ge 0 \forall x\\ |y-2| \ge 0 \forall y \\\end{cases}\\ \text{mà:} |x+3|+|y-2| \le 0\\ \Rightarrow \begin{cases} x+3=0 \rightarrow x = -3\\y-2=0 \rightarrow y = 2\\\end{cases}$
Vậy...........