Giải phương trình sau x^4-4x^3+12x-9=0 x^4-4x^3+3x^2+4x-4=0 x^5-5x^3+4x=0 (x-1)^2-1+x^2=(1-x).(x+3) Giúp mình với ạ mình đang cần gấp

$x^4-4x^3+12x-9=0$

⇔$(x-1)(x^3-3x^2-3x+9)=0$

⇔$(x-1)(x-3)(x^2-3)=0$

⇔$x-1=0; x-3=0;x^2-3=0$

⇔$x=1;x=3;x=√3$

Vậy S={1;3;√3}

$x^4-4x^3+3x^2+4x-4=0$

⇔$(x-1)(x^3-3x^2+4)=0$

⇔$(x-1)(x+1)(x-2)^2=0$

⇔$x-1=0;x+1=0;x-2=0$

⇔$x=1;x=-1;x=2$

Vậy S=[1;-1;2}

$x^5-5x^3+4x=0$

⇔$(x-1)(x^4+x^3-4x^2-4x)=0$

⇔$(x-1)(x+1)(x^3-4x)=0$

⇔$x-1=0;x+1=0;x=0;x-2=0;x+2=0$

⇔$x=1;x=-1;x=0;x=2;x=-2$

Vậy S={1;-1;0;2;-2}

$(x-1)^2-1+x^2=(1-x)(x+3)$

⇔$x^2-2x+1-1+x^2=x+3-x^2-3x$

⇔$2x^2-2x=-2x+3-x^2$

⇔$2x^2-2x+2x-3+x^2=0$

⇔$3x^2-3=0$

⇔$3(x^2-1)=0$

⇔\(\left[ \begin{array}{l}x-1=0\\x+1=0\end{array} \right.\) =>\(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\) 

Vậy S={1;-1}