a) Daca x=6 => (3*6+2)/(6-1) = 20/5 = 4 e Z deci 6 e A
b) x= {-4,0,1,2,6}
daca x = -4 => (3*-4+2)/(-4-1) = -10/-5 = 2 e Z
daca x =0=> (3*0+2)/(0-1) = 2/-1 = -2 e Z
daca x=1 => (3*1+2)/(1-1) = 5/0 = 0 e Z
daca x=2 => (3*2+2)/(2-1)= 8/1 = 8 e Z
daca x = 6 => 4
deci A ={-4,0,1,2,6}
Explicație pas cu pas:
(3x+2)/(x-1) €Z => (x-1)|(3x+2)
Dar cum (x-1)|(x-1)/×3 => (x-1)|3(x-1)
le scadem si avem:
(x-1)|3x+2-3x+3 => (x-1)|5 =>x-1€D5 in Z
=> x-1€{-5,-1,1,5} => x€{-4,0,2,6} => A={-4,0,2,6}
