Cum l-ai scris tu, datorita ordinii operatiilor se intelege ca numai 1se imparte la 6, pt ca impartirea se face inaintea adunarii
deci ar fi asa;
3/2<n+1/6<101/3
9/6<6n+1<202/6
9<36n+6<202
3<36n<196
3/36<n<196/36
1/12<n<49/9
n∈(1/12;49/9)
daca se cere n natural, atunci n∈{1;2;3;4;5}
dacaaaa...era o liniede fractie sub tot (n+1), atunci
era
3/2<(n+1)/6<101/3
si atunci
9/6<(n+1)/6<202/6
9<n+1<202
8<n<201
dac n e real, atunci n∈(8;201)
daca n∈N, atunci n∈{9;10.....200}
