Question

tìm x để A nguyên:

A=\(\dfrac{3}{x+2}\)

Answer 1

`<=> 3 vdots x+2`

`<=> x + 2 in Ư(3)`

`@ x + 2 = 1 => x = -1`

`@ x + 2= -1 => x = -3`

`@ x + 2 = 3 => x = 1`

`@ x + 2 = -3 => x = -5`.

Vậy ...

Answer 2

Để A là số nguyên thì \(x+2\in\left\{1;-1;3;-3\right\}\)

=>\(x\in\left\{-1;-3;1;-5\right\}\)

Answer 3

ĐKXĐ:  \(x+2\ne0\Leftrightarrow x\ne-2\)

Để A nguyên thì \(x+2\inƯ\left(3\right)\)

Mà \(Ư\left(3\right)=\left\{-3;-1;1;3\right\}\)

Khi \(x+2=-3\Leftrightarrow x=-5\)

Khi \(x+2=-1\Leftrightarrow x=-3\)

Khi \(x+2=1\Leftrightarrow x=-1\)

Khi \(x+2=3\Leftrightarrow x=1\)