2x^3+3x^2+3x+1=0

Giải phương trình

Đáp án:

`2x^3+3x^2+3x+1=0`

`⇔ x^3 + 3x^2 + 3x + 1 + x^3 = 0`

`⇔ (x^3 + 3x^2 + 3x + 1) + x^3 = 0`

`⇔ (x + 1)^3 + x^3 = 0`

`⇔ [(x + 1) + x].[(x + 1)^2 - x.(x + 1) + x^2] = 0`

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

`⇔ (2x + 1).(x^2 + x + 1) = 0`

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

`-> x^2 + x + 1 \ne 0`

`-> 2x + 1 = 0`

`⇔ 2x = -1`

`⇔ x = (-1)/2`

Vậy `S = {(-1)/2}`

$#dariana$

$2x^3+3x^2+3x+1=0$

$⇔ 2x^3+2x^2+x^2+2x+x+1=0$

$⇔ (2x^3+2x^2+2x)+(x^2+x+1)=0$

$⇔ 2x(x^2+x+1)+(x^2+x+1)=0$

$⇔ (2x+1)(x^2+x+1)=0$

$⇔$ \(\left[ \begin{array}{l}2x+1=0\\x^2+x+1=0\end{array} \right.\) 

$⇔$ \(\left[ \begin{array}{l}2x=-1\\x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=0\end{array} \right.\) 

$⇔$ \(\left[ \begin{array}{l}2x=-1\\(x+\dfrac{1}{2})^2+\dfrac{3}{4}=0\end{array} \right.\)

$⇔$ \(\left[ \begin{array}{l}x=\dfrac{-1}{2}\\(x+\dfrac{1}{2})^2+\dfrac{3}{4}=0\end{array} \right.\)

$⇔$ \(\left[ \begin{array}{l}x=\dfrac{-1}{2}\\(x+\dfrac{1}{2})^2=\dfrac{-3}{4}\end{array} \right.\)

$⇔$ \(\left[ \begin{array}{l}x=\dfrac{-1}{2}\\x∈∅\end{array} \right.\) vì `(x+\frac{1}{2})^2≥0`

$⇔ x=\dfrac{-1}{2}$

Vậy phương trình có tập nghiệm `S={-1/2}`

$#thanhmaii2008$