Răspuns:
b) [tex]\frac{6}{x+1}[/tex] ∈ Z
x+1 ∈ D₆
x+1 ∈ {1,2,3,6} | -1
x ∈ {0,1,2,5}
c) [tex]\frac{5}{1-x}[/tex] ∈ Z
1-x ∈ D₅
1-x ∈ {1,5} |-1
-x ∈ {0,4} |:(-1)
x ∈ {0,-4}
h) [tex]\frac{-6}{x-2}[/tex] ∈ N
x-2 ∈ D₋₆
x-2 ∈ {-6,-3,-2,-1,1,2,3,6} |+2
x ∈ {-4,-1,0,1,3,4,5,8}
x ∈ N ⇒ x ∈ {0,1,3,4,5,8}
Explicație pas cu pas:
