\(A=\dfrac{2x-5}{x+2}=\dfrac{2x+4-9}{x+2}=\dfrac{2x+4}{x+2}-\dfrac{9}{x+2}\)

Because \(\dfrac{2x+4}{x+2}\) is an integer

=> \(\dfrac{9}{x+2}\) must be an integer.

=> \(x+2\in GCD\left(9\right)\)

=> \(x+2\in\left\{\pm1,\pm3,\pm9\right\}\)

=> \(x\in\left\{-1,-3,1,-5,7,-11\right\}\)

=> \(x\in\left\{\pm1,-3,-5,7,-11\right\}\)

Lê Quốc Trần Anh,you should write \(x+2\in D\left(9\right)\),not \(x+2\in GCD\left(9\right)\)

A = 2 x - 5 x + 2= 2 x + 4 - 9 x + 2= 2 x + 4 x + 2- 9 x + 2A=2X-5X+2=2X+4-9X+2=2X+4X+2-9X+2

2 x + 4 x + 22X+4X+2

9 x + 29X+2

X + 2 ∈ G C D ( 9 )X+2∈GCD(9)

X + 2 ∈ { ± 1 , ± 3 , ± 9 }X+2∈{±1,±3,±9}

X ∈ { - 1 , - 3 , 1 , - 5 , 7 , - 11 }X∈{-1,-3,1,-5,7,-11}

X ∈ { ± 1 , - 3 , - 5 , 7 , - 11 }

\(A=\dfrac{2x-5}{x+2}=\dfrac{2x+4-9}{x+2}=\dfrac{2x+4}{x+2}-\dfrac{9}{x+2}\)

Because \(\dfrac{2x+4}{x+2}\) is an integer

=> \(\dfrac{9}{x+2}\) must be an integer