`\qquad y=x^3+2x^2+(m-3)x+m`
`=>y'=3x^2+4x+m-3`
Ta có:
`y=x^3+2x^2+(m-3)x+m`
`=1/3 x (3x^2+4x+m-3)+2/ 3 x^2+2/ 3 (m-3)x+m`
`=1/ 3x.y' +2/9 . (3x^2+4x+m-3)-8/9x-2/9m+2/3+(2/3 m-2)x+m`
`=1/ 3x.y'+2/9y'+(-8/9+2/3m-2)x-2/9m+m+2/3`
`=y'. (1/3x+2/9)+(2/3m-{26}/9)x+7/9m+2/3`
Vậy `y=y'. (1/3x+2/9)+(2/3m-{26}/9)x+7/9m+2/3`