`A=[(3x-2)(x+1)-(2x+5)(x^2-1)]:(x+1)`
`=[(3x-2)(x+1)-(2x+5)(x+1)(x-1)]/(x+1)`
`={(x+1)[3x-2-(2x+5)(x-1)]}/(x+1)`
`=3x-2-(2x^2+5x-2x-5)`
`=3x-2-2x^2-3x+5`
`=-2x^2+3`
ĐKXĐ : `x \neq -1`
Thay `x=1/2` (tm `x \neq -1`) vào A ta có
`A=-2.(1/2)^2+3`
`=-2. 1/4 + 3`
`=-1/2+6/2`
`=5/2`