Răspunsuri

2014-01-28T17:18:05+02:00
 \frac{x+4}{x+1}  ∈ Z <=>  \frac{x+1+3}{x+1}  ∈ Z => ( \frac{x+1}{x+1} +  \frac{3}{x+1}  ) ∈ Z <=>

<=> 1 +  \frac{3}{x+1}   ∈ Z <=>  \frac{3}{x+1}  ∈ Z ( caci 1 ∈ Z ) =>

=> x + 1 ∈  D_{3} => x+1 ∈ { -3; -1; 1; 3 } => x ∈ { -4; -2; 0; 2 } => 

=> A = { -4; -2; 0; 2 }
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