Đáp án `+` Giải thích các bước giải:

`f) x^4-4x^3+12x-9=0`

`=> x^4-9-4x^3+12x=0`

`=>(x^2)^2 - 3^2 - 4x^3+12x=0`

`=>(x^2-3)(x^2+3)-4x(x^2-3)=0`

`=>(x^2-3)(x^2+3-4x)=0`

`=>(x^2-3)(x^2-x-3x+3)=0`

`=>(x^2-3)x(x-1)-3(x-1)=0`

`=>(x^2-3)(x-3)(x-1)=0`

`=>` $\left[\begin{matrix} x^2-3=0\\ x-3=0 \\x-1=0\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=± \sqrt{3}\\ x=3 \\ x=1\end{matrix}\right.$

Vậy `x ∈ { ±` \sqrt{3} `; 3 ; 1}`.

`h) x^4-4x^3+3x^2+4x-4=0`

`=>x^4-4x^3+4x^2-x^2+4x-4=0`

`=>x^2(x^2-4x+4)-(x^2-4x+4)=0`

`=>(x^2-1)(x^2-4x+4)=0`

`=>(x-1)(x+1)(x-2)^2=0`

`=>`$\left[\begin{matrix} x-1=0\\ x+1=0 \\ (x-2)^2 = 0\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=1\\ x=-1 \\ x=2\end{matrix}\right.$

Vậy `x ∈ {1;-1;2}`.