`a)A= {3x+2}/{x-1} = 3 + 5/{x-1}`

`A ∈ Z ⇔ 5/(x-1) ∈ Z`

`x ∈ Z`

`=> x-1 ∈ Ư(5) = {±1; ±5}`

`=> x ∈ {0; 2; 6; -4}`

`=>x ∈ {0; 2; 6; -4}`

Vậy `3x+2⋮x-1`

`b)x^2+2x-7⋮x+2`

`=> x(x + 2 ) - 7 ⋮ x + 2`

`=> 7 ⋮ x + 2`

`=> x + 2 ∈ {1 ; 7 ; -1 ; - 7}`

`=> x ∈ {-1;5;-3;-9}`

Vậy `x^2+2x-7⋮x+2`

Đáp án:

$a)\,3x+2\ \vdots\ x-1\\\Rightarrow 3x-3+5\ \vdots\ x-1\\\Rightarrow 3x-3\ \vdots\ x-1\Rightarrow 5\ \vdots\ x-1\\\Rightarrow x-1\in Ư(5)\\\Rightarrow x-1\in\{1;-1;5;-5\}\\\Rightarrow x\in\{2;0;6;-4\}\\b)\,x^2+2x-7\ \vdots\ x+2\\\Rightarrow x(x+2)-7\ \vdots\ x+2\\\Rightarrow x(x+2)\ \vdots\ x+2\Rightarrow 7\ \vdots\ x+2\\\Rightarrow x+2\in Ư(7)\\\Rightarrow x+2\in\{1;-1;7;-7\}\\\Rightarrow x\in\{-1;-3;5;-9\}$

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