Question

Determinati valorile intregi ale lui x pentru care :
a)4x +23
supra
2x+1
b)5x+8
supra 2x-1
c)14x +22
supra
2x+1
Toate ecuatiile apartin lui Z

Answer 1

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}