`B = ( 2x^3 - 3x^2 + 5x + 3 )/( 2x - 1 )`

`B = ( 2x^3 - x^2 - 2x^2 + x + 4x - 2 + 5 )/( 2x - 1 )`

`B = ( x^2( 2x - 1 ) - x( 2x - 1 ) + 2( 2x - 1 ) + 5 )/( 2x- 1 )`

`B = x^2 - x + 2 + 5/( 2x - 1 )`

`B in Z => x^2 - x + 2 + 5/( 2x - 1 ) in Z`

`x in Z => 2x - 1 in Z ; 5/( 2x - 1 ) in Z`

`( 2x - 1 ) in Ư( 5 )`

`( 2x - 1 ) in { -5 ; 5 ; 1 ; -1 }`

`2x in { -4 ; 6 ; 2 ; 0 }`

`x in { -2 ; 3 ; 1 ; 0 }` ( thử lại TM )

`Vậy  x in { -2 ; 3 ; 1 ; 0 }` thì `B in Z`