Multimea A
11/2x-1 €Z => (2x-1)€Divizori 11(D indice 11)
D11=1,11
cazu 1
2x-1=1
2x=2
x=1
cazu 2
2x-1=11
2x=12
x=6
Multimea B
3x+5/x+1 €Z=>(x+1)l(3x+5)
(x+1)l(x+1)=>(x+1)l(3x+3) (aici prinzi in acolada ambele relatii =>(x+1)l(3x+5-3x-3)=>(x+1)l2
cazul 1
x+1=1
x=0
cazul 2
x+1=2
x=1
A={1;6}
B={0;1}