{x∈N-{3} I (x-3)/6 < 3/(x-3) < (x-3)/3 }
1. (x-3)/6 < 3/(x-3) inmultim cu 6(x-3)
(x-3)²<18 => (x-3)²∈{0;1;4;9;16}
(x-3)∈{0;1;2;3;4}
x∈{3; 4; 5; 6; 7}
2. 3/(x-3) < (x-3)/3 inmultim cu 3(x-3)
9 < (x-3)² => (x-3)²∈{16; 25; 36; 49; 64;….. }
(x-3)∈{4;5;6;7;8;…. }
x∈{7;8;9;…..}
{3; 4; 5; 6; 7} ∩ {7;8;9;…..}={7}
x=7
