a)

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

`⇔x^3-3x^2+4=0`

`⇔x^3+x^2-4x^2+4=0`

`⇔x^2(x+1)-4(x^2-1)=0`

`⇔x^2(x+1)-4(x+1)(x-1)=0`

`⇔(x+1)(x^2-4(x-1))=0`

`⇔(x+1)(x^2-4x+4)=0`

`⇔(x+1)(x-2)^2=0`

`⇔`\(\left[ \begin{array}{l}x=-1\\x=2\end{array} \right.\) 

b)

`x^4+x^3-4x^2+5x-3=0`

`⇔x^4-x^3+2x^3-2x^2-2x^2+2x+3x-3=0`

`⇔x^3(x-1)+2x^2(x-1)-2x(x-1)+3(x-1)=0`

`⇔(x^3+2x^2-2x+3)(x-1)=0`

`⇔(x^3+3x^2-x^2-3x+x+3)(x-1)=0`

`⇔[x^2(x+3)-x(x+3)+(x+3)](x-1)=0`

`⇔(x^2-x+1)(x+3)(x-1)=0`

Nếu `x^2-x+1=0`

`⇔x^2-2.x. 1/2+1/4+3/4=0`

`⇔(x-1/2)^2=-3/4 (vô lý)`

`⇒x^2-x+1\ne0`

`⇒`\(\left[ \begin{array}{l}x+3=0\\x-1=0\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x=-3\\x=1\end{array} \right.\) 

c)

Cái này có thể sẽ có ý tưởng hay hơn nhưng mình dùng hằng đẳng thức sau nhé:

`(a±b)^4=a^4±4a^3b+6a^2b^2±4ab^3+b^4`

Đặt `t=x-3`

`PT⇒(t-1)^4+(t+1)^4=82`

`⇔t^4-4t^3+6t^2-4t+1+t^4+4t^3+6t^2+4t+1=82`

`⇔2t^4+12t^2+2=82`

`⇔t^4+6t^2-40=0`

`⇔t^4-4t^2+10t^2-40=0`

`⇔t^2(t^2-4)+10(t^2-4)=0`

`⇔(t^2+10)(t^2-4)=0`

`⇔`\(\left[ \begin{array}{l}t^2+10=0\\t^2-4=0\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}t^2=-10(vô lý)\\t^2=4\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}t=2\\t=-2\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x-3=2\\x-3=-2\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x=5\\x=1\end{array} \right.\)