\(A=\dfrac{2x-1}{x+2}=\dfrac{2x+4-5}{x+2}=2-\dfrac{5}{x+2}\)
Để A nguyên thì \(-\dfrac{5}{x+2}\) nguyên
=>\(x+2\inƯ\left(-5\right)\)
=>\(x+2\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{-1;-3;3;-7\right\}\)
\(B=\dfrac{x^2-2x+1}{x+1}=\dfrac{x^2+x-3x-3+4}{x+1}=x-3+\dfrac{4}{x+1}\)
Để B nguyên thì \(\dfrac{4}{x+1}\) nguyên
=>\(x+1\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(x\in\left\{0;-2;1;-3;3;-5\right\}\)
`A = (2x - 1)(x+2) `
Điều kiện: `x ne -2`
`A = (2x + 4 - 5)(x+2) = (2(x + 2) - 5)/(x+2) = 2 - 5/(x+2) `
Để `A ∈ Z` thì: `5 vdots x + 2`
`=> x + 2 ∈ Ư(5) = {-5;-1;1;5}`
`=> x ∈ {-7;-3;-1;3}` (Thỏa mãn)
Vậy ...
`B = (x^2 - 2x + 1)/(x+1)`
Điều kiện: `x ne -1`
`B = (x^2 + 2x + 1 - 4x)/(x+1) = ((x+1)^2 - 4x - 4 + 4)/(x+1) = ((x+1)^2 - 4(x +1) + 4)/(x+1) = x + 1 - 4 + 4/(x+1) `
Để `B ∈ Z` thì: `x+ 1 ∈ Ư(4) = {-4;-2;-1;1;2;4}`
`=> x ∈ {-5;-3;-2;0;1;3}` (Thỏa mãn)
Vậy ...