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

2 câu trả lời

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

`<=>x^4-2x^3 + 3x^3-6x^2 +3x^2-6x +2x-4=0`

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

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

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

`<=>(x-2)[x^2(x+2)+x(x+2)+(x+2)]=0`

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

Do `x^2+x+1=(x+1/2)^2+3/4>= 3/4>0`

`<=>(x-2)(x+2)=0`

`<=>x-2=0` hoặc `x+2=0`

`<=>x=2` hoặc `x=-2`

Vậy `x=2,x=-2`

 

Đáp án:

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

`<=> x^4 + 3x^3 - 2x^3 + 3x^2 - 6x^2 + 2x - 6x - 4 = 0`

`<=> (x^4 + 3x^3 + 3x^2 + 2x ) - ( 2x^3 + 6x^2 + 6x +4) = 0`

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

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

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

Có `x^2 + x +1 = ( x +1/2)^2 + 3/4 > 0`

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

Vậy `S={ ±2}`