Explicație pas cu pas:
4. |2x-7|<9 => -9<2x-7<9 => -2<2x<16 =>
-1<x<8 => x€{1,2,3,4,5,6,7}
2|2x+3|-9<5 => 2|2x+3|<14 => |2x+3|<7 =>
-7<2x+3<7=>-10<2x<4 => -5<x<2 => x€{-4,-3,-2,-1,1}
5. (4n+7)/(2n-1) €Z <=> (2n-1) |(4n+7)
dar stim si ca (2n-1)|(2n-1) => 2n-1|2(2n-1) adica 2n-1|4n-2
Avem 2n-1|4n+7 si 2n-1|4n-2 le scadem si avem
2n-1|4n+7-4n+2 => 2n-1|9 => 2n-1€D9 in Z => 2n-1€{-9,-3,-1,1,3,9} => 2n€{-8,-2,0,2,4,10} => n€{-4,-1,0,1,2,5} => S€{-4,-1,1,2,5}
(7n+13)/(2n-1)€Z <=>(2n-1)|(7n+13)
=>2n-1|2(7n+13)=>2n-1|14n+26
dar cum 2n-1|2n-1 => 2n-1|7(2n-1)=>
2n-1|14n-7
Scadem si avem 2n-1|14n+26-14n+7 =>
2n-1|35 => 2n-1€D35 in Z =>
2n-1€{-35,-7,-5,-1,1,5,7,35} =>
2n€{-34,-6,-4,0,2,6,8,36} =>
n€{-17,-3,-2,0,1,3,4,18} => S€{-17,-3,-2,1,3,4,18}
