Bạn tham khảo nhé.
`3x^3-6x^2+9x=0`
`<=>3x(x^2-2x+3)=0`
`<=>3x(x^2-2x+1+2)=0`
`<=>3x[(x-1)^2+2]=0`
`<=>[(3x=0),((x-1)^2+2=0):}`
`<=>[(x=0),((x-1)^2=-2(\text{Vô lý})):}`
Vậy `x=0`
$\\$
`3x^3-12x^2+12x=0`
`<=>3x(x^2-4x+4)=0`
`<=>3x[x^2-2.2.x+2^2]=0`
`<=>3x(x-2)^2=0`
`<=>[(3x=0),((x-2)^2=0):}`
`<=>[(x=0),(x-2=0):}`
`<=>[(x=0),(x=2):}`
Vậy `x\in{0;2}`