Explicație pas cu pas:
2(3|2x+1|-8)-5< 9 => 2(3|2x+1|-8)<14 =>
3|2x+1|-8<7 => 3|2x+1|<15 => |2x+1|<5 =>
-5<2x+1<5 => -6<2x<4 => -3<x<2 =>
S€{-2,-1,1}
3(2|2x+3|-9)-13<2 => 3(2|2x+3|-9)<15 =>
2|2x+3|-9<5 => 2|2x+3|<14 => |2x+3|<7 =>
-7<2x+3<7 => -10<2x<4 => -5<x<2 =>
S€{-4,-3,-2,-1,1}
|x-1|(|2x-3|-5)<0
Stim ca |x|>=0 adica orice modul este pozitiv deci |x-1|>=0 dar cum avem produs <0 inseamna ca al 2 lea termen este <0 si
|x-1|>0
Deci avem:
|2x-3|-5<0 => |2x-3|<5 => -5<2x-3<5 =>
-2<2x<8 => -1<x<4 =>S€{1,2,3}
|x+2|(7-|2x+5|)>0
Avem ca la cel anterior |x+2|>0 deci celalalt termen este >0
7-|2x+5|>0=> |2x+5|<7 => -7<2x+5<7 =>
-12<2x<2 => -6<x<1 => S€{-5,-4,-3,-2,-1}
