Question

9≥|12-(х+3)|>-1
Решите, пожалуйста, системой неравенств! СРОЧНО!!! ОЧЕНЬ НУЖНО!!!!!

Answer 1

$9 \geq |12-(x+3)| \geq- 1\\-1 \leq |12-x-3| \leq 9\\-1 \leq |9-x| \leq 9\\-1 \leq |x-9| \leq 9\\\\ \left \{ {{|x-9| \geq -1} \atop {|x-9| \leq 9}} \right. =\ \textgreater \ \left \{ {{x\in R} \atop {-9 \leq x-9 \leq 9}} \right. =\ \textgreater \ \left \{ {x\in R} \atop {-9+9\leq x \leq 9+9}} \right. =\ \textgreater \ \left \{ {{x \in R} \atop {0 \leq x \leq 18}} \right.\\\\x\in[0;18]$