Bài `2:`
`(x^4 - 3x^2 + 9)(x^2 + 3) - (3 + x^2)^3`
` = (x^2 + 3) . [ (x^4 - 3x^2 + 9) - (x^2 + 3)^2]`
` = (x^2 + 3) . [ x^4 - 3x^2 + 9 - (x^4 +6x^2+9)]`
` = (x^2 + 3) . (x^4 - 3x^2 + 9 - x^4 - 6x^2 - 9)`
` = (x^2 + 3). (-9x^2)`
` = -9x^4 - 27x^2`
Bài `3:`
`(3+x)^3 - x . (3x+1)^2 + (2x+1).(4x^2 - 2x+1) = 28`
`=> (x^3 + 9x^2 + 27x+27) - x . (9x^2 + 6x+1) + (2x)^3 + 1^3 = 28`
`=> x^3 + 9x^2 + 27x + 27 - 9x^3 - 6x^2- x + 8x^3 + 1 =28`
`=> (x^3 - 9x^3 + 8x^3) + (9x^2 - 6x^2) + (27x -x) + (27+1) = 28`
`=> 3x^2 + 26x + 28 = 28`
`=> 3x^2 + 26x + 28 -28=0`
`=> 3x^2 + 26x = 0`
`=> 3x . (x + 26/3) = 0`
`=> 3x=0` hoặc `x+26/3=0`
`+)3x=0 =>x=0`
`+)x+26/3=0=>x=-26/3`
Vậy `x\in{ 0; -26/3}`
