$a)x^3+4x^2+x-6=0$ $b)x^3-6x^2+11x-6=0$ $c)x^3-4x^2+x+6=0$ $d)x^3-3x^2+4=0$

Đáp án + Giải thích các bước giải:

`a, x^3+4x^2+x-6=0`

`⇔ x^3+2x^2+2x^2+4x-3x-6=0`

`⇔ x^2(x+2)+2x(x+2)-3(x+2)=0`

`⇔ (x+2)(x^2+2x-3)=0`

`⇔ (x+2)(x^2-x+3x-3)=0`

`⇔ (x+2)[x(x-1)+3(x-1)]=0`

`⇔ (x+2)(x-1)(x+3)=0`

`⇔ [(x=-2),(x=1),(x=-3):}`

Vậy `S={-2;1;-3}`

`b, x^3-6x^2+11x-6=0`

`⇔ x^3-x^2-5x^2+5x+6x-6=0`

`⇔ x^2(x-1)-5x(x-1)+6(x-1)=0`

`⇔ (x-1)(x^2-5x+6)=0`

`⇔ (x-1)(x^2-2x-3x+6)=0`

`⇔ (x-1)[x(x-2)-3(x-2)]=0`

`⇔ (x-1)(x-2)(x-3)=0`

`⇔ [(x=1),(x=2),(x=3):}`

Vậy `S={1;2;3}`

`c, x^3-4x^2+x+6=0`

`⇔ x^3+x^2-5x^2-5x+6x+6=0`

`⇔ x^2(x+1)-5x(x+1)+6(x+1)=0`

`⇔ (x+1)(x^2-5x+6)=0`

`⇔ (x+1)(x^2-2x-3x+6)=0`

`⇔ (x+1)[x(x-2)-3(x-2)]=0`

`⇔ (x+1)(x-2)(x-3)=0`

`⇔ [(x=-1),(x=2),(x=3):}`

Vậy `S={-1;2;3}`

`d, x^3-3x^2+4=0`

`⇔ x^3-2x^2-x^2+4=0`

`⇔ x^2(x-2)-(x^2-2^2)=0`

`⇔ x^2(x-2)-(x-2)(x+2)=0`

`⇔ (x-2)(x^2-x-2)=0`

`⇔ (x-2)(x^2+x-2x-2)=0`

`⇔ (x-2)[x(x+1)-2(x+1)]=0`

`⇔ (x-2)(x+1)(x-2)=0`

`⇔ [(x=2),(x=-1):}`

Vậy `S={2;-1}`