a) (4x + 23)/(2x + 1) ∈ Z ⇔ (4x + 23) si (2x+ 1) au un divizor comun ≠ 1
daca d | (4x + 23) (1)
si d | (2x + 1) ⇒ d | 2(2x+1) = 4x + 2 (2)
⇒ d | [(1) - (2)] = 21 ⇒ d∈{ - 21; - 7; - 3; -1; 1; 3; 7 ;21}
ptr. 2x + 1 = -21 x = -11 (4x + 23)/(2x+1) = -21/-21 = 1 ∈Z
2x+1 = -7 x = -4
2x + 1 = -3 x = -2
2x +1 = -1 x = -1
2x + 1 = 1 x = 0
2x + 1 = 3 x = 1
2x +1 = 7 x = 3
2x+1 = 21 x= 10
b) (5x+8)/(2x-1) ∈ Z⇒ (2x-1) | (5x +8)
(2x - 1) | (2x -1) ⇒ (2x-1) | 5(2x-1) = 10x - 5 (1)
(2x-1) | (5x+8) ⇒ (2x-1) | 2(5x+8) = 10x + 16 (2)
(2x-1) | [(2)-(1)] = 21 ⇒ ( 2x- 1) ∈ {-21,-7,-3,-1,1,3,7,21]
2x∈{-20,-6,-2,0,2,4,8,22} ⇒ x ∈ { -10,-3, -1, 0, 1, 2, 4,11}
c) (14x+22)/(2x + 1) ∈ Z ⇔ (2x+1) | (14x + 22)
(2x + 1) | (2x+1) ⇒ (2x+1) | 7(2x +1) = 14x + 7 (1)
(2x+1) | (14x +22) (2)
⇒ (2x+1) | [(2) - (1)] = 15 ⇒⇒ (2x+1) ∈ { -15, -5, - 3, -1, 1, 3, 5, 15}
⇒ 2x ∈{ -16, - 6, -4, -2, 0, 2, 4, 14}
⇒ x ∈ { -8, -3, -2, -1, 0, 1, 2, 7}
